assignment 11.2 physics class 12

Exercise 11.2 Physics Class 12

Ncert Exercise 11.2 Physics Class 12 Pdf and Video Solution

Here you will get the pdf solution for Each and Every question of this chapter of NCERT

physics chapter 11 class 12 ncert solutions  (100% working)

Click here (You will be redirected to the next page)

And For Video solution of this question Stay on this page

Ncert Exercise 11.2 Physics Class 12

NCERT Solutions for Class 12 Physics Chapter 11 Free PDF Download

Ncert solutions for class 12 physics chapter 11 dual nature of radiation and matter.

Physics is a subject that studies the universe and how it works. In addition, it derives formulas on the basis of their functioning. But class 12 physics is quite hard to understand for students. That’s why we prepared NCERT Solutions for Class 12 Physics Chapter 11 which will direct students through the chapter in an easy manner. Also, our video tutorial explains each topic in simple language with example.

These solutions are pretty easy to understand as our expert teacher has designed them to convey a proper and complete meaning of each topic and formulas of the chapter.

Toppr provides free study materials, previous years question papers, and 1000+ hour of video lectures for  free . Download Toppr app for  Android  and  iOS  or  signup  for free.

Download NCERT Solutions for Class 12 Physics here

Download NCERT Solutions for Class 12 here

Download NCERT Solutions for all subjects here

CBSE Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter NCERT Solutions

To introduce, dual nature of radiation refers to the wave nature and particle nature of the electrons. While dual nature of matter means that the matter shows duel properties of particle and wave nature. Moreover, in the NCERT Solutions for Class 12 Physics Chapter 11 consist of all the study material or guides that students require to score decent marks in exams.

Subtopics covered under NCERT Solutions for Class 12 Physics Chapter 11

11.1 introduction.

This topic explains what different physicist says about the dual nature of different elements. Also, the topic describes what is the dual nature of radiation and matter.

11.2 Electron Emission

This topic defines how an electron escapes from a metal surface. Also, the topic explains the work function of some elements in the form of a chart.

11.3 Photoelectric Effect

11.3.1 Hertz’s Observation- This topic defines when and during which experiment Heinrich Hertz discovers the photoelectric effect.

11.3.2 Hallwach’s and Lenard’s Observation- This topic defines what they discovered about photoelectric effect during their experiment.

11.4 Experimental Study of Photoelectric Effect

This topic outlines the experiment that reveals the various properties and effects of the photoelectric effect.

11.4.1 Effect on the intensity of light on the photocurrent- This topic describes the effect of photoelectric current effect in the presence of light.

11.4.2 Effect of potential on photoelectric current- This topic overviews the positive and negative effects on photoelectric current.

11.4.3 Effect of Frequency of Incident Radiation on stopping Potential- This topic defines how incident radiation affects the potential of photoelectric current. Moreover, the topic contains a graph, which shows the potential of two elements. And, on the basis of this certain observations are made.

11.5 photoelectric Effect and Wave Theory of Light

This topic explains the photoelectric effect in the presence of wave theory of light (the light is an electromagnetic ray that extends over a region). Also, how it affects the functioning of the photoelectric effect.

11.6 Einstein’s Photoelectric Equation: Energy Quantum of Radiation

This topic defines how Einstein experimented on photoelectric current and after some observation derives a formula out of it.

11.7 Particle Nature of Light: The Photon

This topic defines what is a photon, how it works are its properties.

11.8 Wave Nature of Matter

This topic describes the wave nature of objects in dual wave-particle. Besides, the topic has some experiments, and formula related to it.

11.9 Davisson and Germer Experiment

This topic outlines the experiment that these two conducted to find the wave nature of the electron. Also, what they observe there during their experiment.

Summary- It sums up all the topics in pointers. Also, it mentions each formula of the chapter.

You can download NCERT Solutions for Class 12 Physics Chapter 11 PDF by clicking on the button below.

Download Toppr – Best Learning App for Class 5 to 12

Toppr is the best leaning platform in India that has all the NCERT Solutions from Class 5 to 12. Besides, it brings the easiest and convenient solution for students. Furthermore, it facilitates a competitive environment for students to score good marks. In addition, it provides some other benefits like video tutorials, mock test live classes and many more. To learn fast download the Toppr Android or iOS app or Signup for free.

Customize your course in 30 seconds

Which class are you in.

NCERT Solutions for Class 12

• NCERT Solutions for Class 12 Chemistry Chapter 4 Free PDF Download
• NCERT Solutions for Class 12 Chemistry Chapter 15 Free PDF Download
• NCERT Solutions for Class 12 Biology Chapter 5 Free PDF Download
• NCERT Solutions for Class 12 Biology Chapter 7 Free PDF Download
• NCERT Solutions for Class 12 Biology Chapter 6 Free PDF Download
• NCERT Solutions for Class 12 Biology Chapter 4 Free PDF Download
• NCERT Solutions for Class 12 Biology Chapter 3 Free PDF Download
• NCERT Solutions for Class 12 Biology Chapter 9 Free PDF Download
• NCERT Solutions for Class 6 Social Science Chapter 12 Free PDF Download
• NCERT Solutions for Class 12 Chemistry Chapter 6 Free PDF Download

One response to “NCERT Solutions for Class 12 Chemistry Chapter 6 Free PDF Download”

Can you convert it in Hindi

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Download the App

Talk to our experts

1800-120-456-456

• NCERT Exemplar for Class 12 Physics Chapter-11 (Book Solutions)
• Textbook Solutions

NCERT Exemplar for Class 12 Physics - Dual Nature Of Radiation And Matter - Free PDF Download

Free PDF download of NCERT Exemplar for Class 12 Physics Chapter 11 - Dual Nature Of Radiation And Matter solved by expert Physics teachers on Vedantu.com as per NCERT (CBSE) Book guidelines. All Chapter 11 - Dual Nature Of Radiation And Matter exercise questions with solutions to help you to revise complete syllabus and score more marks in your examinations.

Access NCERT Exemplar Solutions for CBSE Class 12 Science (Physics) Chapter 11 - Dual Nature of Radiation and Matter

Multiple choice questions-1.

1. A particle is dropped from a height $H$. The de Broglie wavelength of the particle as a function of height is proportional to 

(b) ${H^{\dfrac{1}{2}}}$

 (c) ${H^0}$

(d) ${H^{ - \;\dfrac{1}{2}}}$

Ans: When the body is falling freely from a height, the velocity is

\[ = \sqrt {2gH} \]

according to de-Broglie wavelength 

$\lambda  = \dfrac{h}{p}$

$\lambda  = \dfrac{h}{{mv}}$

$ = \dfrac{h}{{m\sqrt {2gH} }}$

Where, 

$h$, $m$ and $g$ are constant.

$\therefore \dfrac{h}{{m\sqrt {2g} }}$is constant

$ \Rightarrow \lambda  \propto \dfrac{1}{{\sqrt H }}$

$ \Rightarrow \lambda  \propto {H^{ - \,\dfrac{1}{2}}}$

Thus, option (d) is correct.

2. The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with $1\;MeV$ energy is nearly  

(a) $1.2\,nm$

(b) $1.2 \times {10^{ - 3}}\,nm$

(c) $1.2 \times {10^{ - 6}}\;nm$

(d) $1.2 \times {10^1}\,nm$

Ans:   we know that the energy of the photon should be equal to the binding energy of proton

Thus, energy of photon $ = 1\;MeV$

$ = {10^6} \times 1.6 \times {10^{ - 19}}J$

$\lambda  = \dfrac{{hc}}{E}$

$ \Rightarrow \dfrac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.6 \times {{10}^{ - 13}}}}$

$ \Rightarrow \dfrac{{6.63 \times 3}}{{1.60}} \times {10^{ - 26 + 13}}$

$ \Rightarrow \dfrac{{19.89}}{{1.60}} \times {10^{ - 13}}$

$ = 12.4 \times {10^1} \times {10^{ - 13}}$

$ \Rightarrow 1.24 \times {10^{ - 9}} \times {10^{ - 3}}$

$ = 1.24 \times {10^{ - 3}}\;nm$

So, option (b) is correct.

3. Consider a beam of electrons (each electron with energy ${E_0}$ ) incident on a metal surface kept in an evacuated chamber. Then 

(a) no electrons will be emitted as only photons can emit electrons.

(b) electrons can be emitted but all with an energy, ${E_0}$.

(c) electrons can be emitted with any energy, with a maximum of ${E_0}$ – $\phi $($\phi $ is the work function).

(d) electrons can be emitted with any energy, with a maximum of ${E_0}$ .

Ans:   When a beam of electrons in each electron of energy ${E_0}$ incident on the metal surface which is kept in vacuum then because of elastic collisions with electrons on the surface, the energy of incident electrons will be transferred to the emitted electrons. For the emission of electrons below the surface the part of ${E_0}$ of incident electrons is consumed against work function so the energy of emitted electrons becomes less than ${E_0}$. Thus, the maximum energy of emitted electrons can be ${E_0}$. So, the option (d) is correct.

4. Consider \[{{Fig}}.{\text{ }}{{11}}.{{7}}\] in the NCERT text book of physics for Class XII. Suppose the voltage applied to \[{{A}}\] is increased. The diffracted beam will have the maximum at a value of \[{{\theta }}\] that 

(a) will be larger than the earlier value. 

(b) will be the same as the earlier value. 

(c) will be less than the earlier value. 

(d) will depend on the target

Ans:  

Through Davisson-Germer experiment, we know that diffracted beam of electrons

has the de-Broglie wavelength

${\lambda _d} = \dfrac{{12.27}}{{\sqrt V }}\mathop A\limits^o $ ….(i) 

Here, $V$ is the voltage applied. When there is a maximum of the diffracted electrons at an angle $\theta $, then 

$2d\sin \theta  = \lambda $

From equation (i), as there are increases in the applied voltage in this experiment, the wavelength ${\lambda _d}$ decreases in turn $\sin \theta $ or $\theta $ decreases by relation (ii). Thus, this verifies the option (c).

5. A proton, a neutron, an electron and an α-particle have the same energy. Then their de Broglie wavelengths compare as 

(a) ${\lambda _p} = {\lambda _n} > {\lambda _e} > {\lambda _\alpha }$

(b) ${\lambda _\alpha } < {\lambda _p} = {\lambda _n} > {\lambda _e}$

(c) ${\lambda _e} < {\lambda _p} = {\lambda _n} > {\lambda _\alpha }$

(d) ${\lambda _e} = {\lambda _p} = {\lambda _n} = {\lambda _\alpha }$

Ans:   According to de-Broglie wavelength

${\lambda _d} = \dfrac{h}{p}$

${E_p} = {E_n} = {E_e} = {E_\alpha }$

$K.E. = K = \dfrac{1}{2}m{v^2}$

$ \Rightarrow 2K = m{v^2}$

$ \Rightarrow 2Km = {m^2}{v^2}$

$ \Rightarrow 2mK = {p^2}$

$ \Rightarrow \sqrt {2mK}  = p$

$\therefore {\lambda _d} = \dfrac{h}{p}$

$ \Rightarrow {\lambda _d} = \dfrac{h}{{\sqrt {2mK} }}$

Or ${\lambda _d} = \dfrac{h}{{\sqrt {2mK} }}$

Where, $h$ and $E$$\left( {K.E.} \right)$ is constant.

$\therefore \lambda  \propto \dfrac{1}{{\sqrt m }}$

Thus, 

${m_\alpha } > {m_p} = {m_n} > {m_e}$

So, ${\lambda _\alpha } < {\lambda _p} = {\lambda _n} > {\lambda _e}$, option (b) is correct.

6. An electron is moving with an initial velocity $v = {v_0}\hat i$ and is in a magnetic field $B = {B_0}\hat j$. Then it’s de Broglie wavelength 

(a) remains constant. 

(b) increases with time. 

(c) decreases with time. 

(d) increases and decreases periodically.

Ans:   In the question, it is given that

$v = {v_0}\hat i$

And, $B = {B_0}\hat j$

The force on moving electron is perpendicular magnetic field $B$ is,

$F =  - e\left( {\vec v = \vec B} \right)$

$ =  - e\left[ {{v_0}\hat i \times {B_0}\hat j} \right]$

$ =  - e{v_0}{B_0}\hat i \times \hat j$

$F =  - e{v_0}{B_0}\hat k$

So, here the force is perpendicular to both $v$ and $B$ as it is perpendicular to the velocity. Thus, there will be no change in $v$ or $mv$ so the de-Broglie wavelength remains the same. 

7. An electron (mass $m$) with an initial velocity $v = {v_0}\hat i\left( {{v_0} > 0} \right)$ is in an electric field $E =  - {E_0}\hat i$(${E_0} = {\text{ }}{{constant}}{\text{ }} > {\text{ }}{{0}}$). It’s de-Broglie wavelength at time \[{{t}}\] is given by 

(a) $\dfrac{\lambda }{{\left[ {1 + \dfrac{{e{E_0}}}{m}\dfrac{t}{{{v_0}}}} \right]}}$

(b) ${\lambda _0}\left[ {1 + \dfrac{{e{E_0}}}{{m{v_0}}}t} \right]$

(c) ${\lambda _0}$

(d) ${\lambda _0}t$

Ans:   at initial, de-Broglie wavelength 

${\lambda _0} = \dfrac{h}{{m{v_0}}}$

Force acting on electron

$F = \left( { - e} \right)\left( { - {E_0}i} \right)$

$ma = e{E_0}\hat i$

$a = \dfrac{{e{E_0}}}{m}\hat i$

After time $t$, velocity of electron is

$v = {v_0} + at$

$v = {v_0}i + \dfrac{{e{E_0}}}{m}i \cdot t$

$v = \left[ {{v_0} + \dfrac{{e{E_0}t}}{m}} \right]\hat i$

$\therefore $New de-Broglie wavelength $\lambda  = \dfrac{h}{{mv}}$

$\lambda  = \dfrac{h}{{m\left[ {{v_0} + \dfrac{{e{E_0}t}}{m}} \right]i}}$

$ = \dfrac{h}{{m{v_0}\left[ {1 + \dfrac{{e{E_0}t}}{{m{v_0}}}} \right]}}$

$\lambda  = \dfrac{{{\lambda _0}}}{{\left[ {1 + \dfrac{{e{E_0}t}}{{m{v_0}}}} \right]}}$$\left( {\because \dfrac{h}{{m{v_0}}} = {\lambda _0}\,\,from\,\,eqn.I} \right)$

8. An electron (mass\[{{m}}\]) with an initial velocity $v = {v_0}\hat i$ is in an electric field $E = {E_0}\hat j$ . If, ${\lambda _0} = \dfrac{h}{{m{v_0}}}$, it’s de-Broglie wavelength at time \[{{t}}\]is given by  

(a) ${\lambda _0}$

(b) ${\lambda _0}\sqrt {1 + \dfrac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} $

(c) $\dfrac{{{\lambda _0}}}{{\sqrt {1 + \dfrac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} }}$

(d) $\dfrac{{{\lambda _0}}}{{\left[ {1 + \dfrac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} \right]}}$

Ans:   We know that de-Broglie wavelength 

Force acting on the moving electron due to electric field $E$

$F =  - eE$

$ \Rightarrow F =  - e{E_0}\hat j$

Because of electric field, acceleration in electron due to force,

$ma =  - e{E_0}\hat j$ 

$ \Rightarrow a = \dfrac{{ - e{E_0}}}{m}\hat j$

And acceleration on electron acts along $ - Y$direction, thus the initial velocity of electron along $X - $axis

${v_{{x_0}}} = {v_0}\hat i$

Thus, the initial velocity of electron in $Y$direction is zero

$\therefore {v_{{y_0}}} = 0$

And the velocity of electron after time $t$ along $X - $axis

${v_x} = {v_0}\hat i$

Thus, velocity of electron after $t$along $Y - $axis

$v = u + at$

${v_y} = 0 + \left( {\dfrac{{ - e{E_0}}}{m}\hat j} \right)t$

${v_y} = \dfrac{{ - e{E_0}}}{m}\hat jt$

After time $t$, magnitude of velocity of electron is

$v = \sqrt {v_x^2 + v_y^2} $

$v = \sqrt {v_0^2 + {{\left( {\dfrac{{ - e{E_0}}}{m}\hat jt} \right)}^2}} $

$\left| v \right| = {v_0}\sqrt {1 + {e^2}E_0^2{t^2}} $

$\therefore \lambda ' = \dfrac{h}{{mv}}$

$ = \dfrac{h}{{m{v_0}\sqrt {1 + \dfrac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} }}$

$v' = \dfrac{1}{{\sqrt {1 + \dfrac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}\left( {\because {\lambda _0} = \dfrac{h}{{m{v_0}}}} \right)} }}$

 Thus, option (c) is verified.

MULTIPLE CHOICE

Questions-ii more than.

9. Relativistic corrections become necessary when the expression for the kinetic energy$\dfrac{1}{2}m{v^2}$ , becomes comparable with $m{c^2}$ , where \[{{m}}\]is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron? 

(a) \[{{\lambda }}{\text{ }} = {{10}}\,{{nm}}\]

(b) \[{{\lambda }}{\text{ }} = {{1}}{{{0}}^{--{{1}}}}{{nm}}\]

(c) \[{{\lambda }}{\text{ }} = {{1}}{{{0}}^{--{{4}}}}{{nm}}\]

(d) \[{{\lambda }}{\text{ }} = {{1}}{{{0}}^{--{{6}}}}{{nm}}\]

Ans:   according to de-Broglie wavelength 

$ \Rightarrow v = \dfrac{h}{{m\lambda }}$

Here, $h = 6.6 \times {10^{ - 34}}$

$m = 9 \times {10^{ - 31}}\,kg$

$ = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{9 \times {{10}^{ - 31}}\lambda }}$

$ = \dfrac{{6.6 \times {{10}^{ - 34 + 31}}}}{{9\lambda }}$

$ = \dfrac{{0.73 \times {{10}^{ - 3}}}}{\lambda }$

$(a)$ $\lambda  = 10\;nm$

$ = 10 \times {10^{ - 9}}\;m$

$ = {10^{ - 8}}\;m$

$\therefore v = \dfrac{{7.3 \times {{10}^{ - 4}}}}{{{{10}^{ - 8}}}}$

$ = 7.3 \times {10^4}$

$ \approx {10^4} < 3 \times {10^8}$(Speed of light)

$\left( b \right)$$\lambda  = {10^{ - 1}}\;nm$

$ = {10^{ - 1}} \times {10^{ - 9}}\;m$

$ = {10^{ - 10}}\;m$

$\therefore v = \dfrac{{7.3 \times {{10}^{ - 4}}}}{{{{10}^{ - 10}}}}$

$ = 7.3 \times {10^{ - 4 + 10}}$

$ = 7.3 \times {10^6}$

$ \approx {10^7} < {10^8}$(Speed of light)

(c) $\lambda  = {10^{ - 4}}\;nm$

$ = {10^{ - 4}} \times {10^{ - 9}}\;m$

$ = {10^{ - 13}}\;m$

$\therefore v = \dfrac{{7.3 \times {{10}^{ - 4}}}}{{{{10}^{ - 13}}}}$

$ = 7.3 \times {10^{ - 4 + 13}}$

$7.3 \times {10^9}$

$ \approx {10^9} > {10^8}$(Speed of light)

(d) $\lambda  = {10^{ - 6}}\;nm$

$ = {10^{ - 6}} \times {10^{ - 9}}\;m$

$ = {10^{ - 15}}\;m$

$\therefore v = \dfrac{{7.3 \times {{10}^{ - 4}}}}{{{{10}^{ - 15}}}}$

$ = 7.3 \times {10^{11}}\,m$

$ \approx {10^{11}} > {10^8}$(Speed of light)

Here we can see that the velocity of electron is more for option (c) and (d) where the relativistic correction become necessary although the speed of electron is $7.3 \times {10^6}\,m{s^{ - 1}}$ which is comparable with the speed of light, thus the correct answers of this question is option (c) and (d).

10. Two particles\[\;{{{A}}_{{1}}}\]and\[\;{{{A}}_{{2}}}\] of masses\[{{{m}}_{{1}}}\], \[{{{m}}_{{2}}}\](\[{{{m}}_{{1}}}{\text{ }} > {\text{ }}{{{m}}_{{2}}}\]) have the same de-Broglie wavelength. Then 

(a) their momenta are the same. 

(b) their energies are the same. 

(c) energy of \[\;{{{A}}_{{1}}}\] is less than the energy of \[\;{{{A}}_{{2}}}\]. 

(d) energy of \[\;{{{A}}_{{1}}}\] is more than the energy of \[\;{{{A}}_{{2}}}\].

Ans:  We know that de-Broglie wavelength

$p = \dfrac{h}{\lambda }$

or $p \propto \dfrac{1}{\lambda }$

$\dfrac{{{p_1}}}{{{p_2}}} = \dfrac{{{\lambda _2}}}{{{\lambda _1}}}$

$\because {\lambda _1} = {\lambda _2} = \lambda $(given)

${p_1} = {p_2}$this verifies option (a)

${E_a} = \dfrac{1}{2}m{v^2}$

$ = \dfrac{1}{2}\dfrac{{{m^2}{v^2}}}{m}$

$ = \dfrac{{{p^2}}}{{2m}}$

$E \propto \dfrac{1}{m}$$\left[ {as\;{p_1} = {p_2} = constant\left( {as\;proved\;above} \right)} \right]$

$\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{m_2}}}{{{m_1}}}$

$\dfrac{{{E_1}}}{{{E_2}}} < 1$$\left[ {\because {m_1} > {m_2}\left( {given} \right)} \right]$

$\therefore {E_2} > {E_1}$ this verifies answer (c).

11. The de Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron. The speed of the electron is ${v_e} = \dfrac{c}{{100}}$ Then 

(a) $\dfrac{{{E_e}}}{{{E_p}}} = {10^{ - 4}}$

(b) $\dfrac{{{E_e}}}{{{E_p}}} = {10^{ - 2}}$

(c) $\dfrac{{{p_e}}}{{{m_e}c}} = {10^{ - 2}}$

(d) $\dfrac{{{p_e}}}{{{m_e}c}} = {10^{ - 4}}$

Ans:   de-Broglie wavelength 

$\therefore {\lambda _e} = \dfrac{h}{{{m_e}{v_e}}}$

$ = \dfrac{h}{{{m_e}\dfrac{c}{{100}}}}$

$ = \dfrac{{100h}}{{{m_e}c}}$  ….(i)

$K.E. = \dfrac{1}{2}m{v^2}$

$ \Rightarrow E = \dfrac{{{m^2}{v^2}}}{{2m}}$   ….(ii)

$ \Rightarrow {m_e}{v_e} = \sqrt {2{m_e}{E_e}} $

${\lambda _e} = \dfrac{h}{{{m_e}{v_e}}}$

$ = \dfrac{h}{{\sqrt {2{m_e}{E_e}} }}$ …[from (ii)]

$ \Rightarrow {E_e} = \dfrac{{{h^2}}}{{2{m_e}\lambda _e^2}}$ ….(iii)

For proton 

${\lambda _p} = 2{\lambda _e}$ [given]

$\therefore {E_p} = \dfrac{{hc}}{{{\lambda _p}}}$

$ = \dfrac{{hc}}{{2{\lambda _e}}}$

Now, $\dfrac{{{E_p}}}{{{E_e}}} = \dfrac{{hc}}{{2{\lambda _e}}} \times \dfrac{{2{m_e}\lambda _e^2}}{{{h^2}}}$

$ = \dfrac{{{\lambda _e}{m_e}c}}{h}$[from (iii)]

$ = \dfrac{{100h}}{{{m_e}c}} \times \dfrac{{{m_e}c}}{h}$

$ \Rightarrow \dfrac{{{E_e}}}{{{E_p}}} = {10^{ - 2}}$ this verifies option (b)

Now, ${p_e} = {m_e}{v_e}$

$ = {m_e} \times \dfrac{c}{{100}}$$\left( {\because {v_e} = \dfrac{c}{{100}}} \right)$

$ \Rightarrow \dfrac{{{p_e}}}{{{m_e}c}} = \dfrac{1}{{100}}$

$ = {10^{ - 2}}$ this verifies option (c).

12. Photons absorbed in matter are converted to heat. A source emitting \[{{n}}{\text{ }}{{photon}}\;{{se}}{{{c}}^{ - 1}}\]of frequency \[{{\nu }}\]is used to convert \[{{1}}\,{{kg}}\] of ice at \[{{0}}^\circ {{C}}\] to water at\[{{0}}^\circ {{C}}\]. Then, the time\[\;{{T}}\] taken for the conversion 

(a) decreases with increasing\[{{n}}\], with\[\;{{\nu }}\] fixed. 

(b) decreases with \[{{n}}\] fixed, \[\;{{\nu }}\] increasing 

(c) remains constant with \[{{n}}\] and \[\;{{\nu }}\] changing such that \[nv = \] constant. 

(d) increases when the product \[nv\]increases.

Ans:   the amount of heat required to convert $1\;kg$ of ice at ${0^\circ }C$to water at${0^\circ }C$ is

Where, $n = $no. of photons incident per second.

Let time $T$ be taken by radiation to melt the ice at ${0^\circ }C$. Then,

$E = nhv\,t$

$mL = n\,\,hv\,t$

$t = \dfrac{{mL}}{{nh\,v}}$

$\therefore t \propto \dfrac{1}{n}$and $T \propto \dfrac{1}{v}$

Or $t \propto \dfrac{1}{{nv}}$[$\because m$,$L$and $h$ are constant]

This proves option (a), (b) and (c).

$\therefore $if $nv$ increases, $T$decreases which does not verify option (d).

13. A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de-Broglie wavelength of the particle varies cyclically between two values \[{{{\lambda }}_{{1}}}\]and \[{{{\lambda }}_{{2}}}\]with \[{{{\lambda }}_{{1}}} > {{{\lambda }}_{{2}}}\]. Which of the following statements are true? 

(a) The particle could be moving in a circular orbit with origin as centre 

(b) The particle could be moving in an elliptic orbit with origin as its focus. 

(c) When the de Broglie wavelength is \[{{{\lambda }}_{{1}}}\], the particle is nearer the origin than when its value is\[{{{\lambda }}_{{2}}}\]

(d) When the de-Broglie wavelength is \[{{{\lambda }}_{{2}}}\], the particle is nearer the origin than when its value is \[{{{\lambda }}_{{1}}}\]. 

Ans:   The de-Broglie wavelength varies regularly between two values ${\lambda _1}$ and ${\lambda _2}$ of the particle. This is possible when the particle is operating in an elliptical orbit with origin as one of its focuses then. As if ${\lambda _1}$ and ${\lambda _2}$ are equal, then their speed must be equal and particle must move in circular orbit. Hence this verifies option (b).

Let ${v_1}$ and ${v_2}$ be the speeds of particles at $A$ and $B$ respectively and origin is at $O$. If particle $A$ and $B$ associated with the de-Broglie wavelengths ${\lambda _1}$ and ${\lambda _2}$ respectively, then

${\lambda _1} = \dfrac{h}{{m{v_1}}}$

And ${\lambda _2} = \dfrac{h}{{m{v_2}}}$ 

So, $\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{{v_2}}}{{{v_1}}}$$\left[ {\because {\lambda _1} > {\lambda _2}} \right]$

Thus, ${v_1} < {v_2}$  …..(i)

According to the law of conservation of angular momentum, the speed of the particle will be more when it is closer to focus. This proved the option (d).

VERY SHORT ANSWER TYPE

14. A proton and an α-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths \[{{{\lambda }}_{{p}}}\]and \[{{{\lambda }}_\alpha }\]related to each other?

Ans:   The both particles are promoted at the same potential difference so their $K.E.s$ will be equal i.e.,

${K_1} = {K_2} = K = qV$$\left( {\because V = \dfrac{W}{q}} \right)$

So, $\lambda  = \dfrac{h}{{\sqrt {2mK} }}$

${\lambda _d} = \dfrac{h}{{\sqrt {2mqV} }}$

$\therefore \dfrac{{{\lambda _p}}}{{{\lambda _\alpha }}} = \dfrac{h}{{\sqrt {2{m_p}{q_p}{V_p}} }} \times \dfrac{{\sqrt {2{m_\alpha }{q_\alpha }{V_\alpha }} }}{h}$

${m_\alpha } = 4\,{m_p}$; ${q_\alpha } = 2e$; ${q_p} = e$;${V_p} = {V_\alpha } = V$(P.D. applied)

$\dfrac{{{\lambda _p}}}{{{\lambda _\alpha }}} = \sqrt {\dfrac{{2 \times 4{m_p}2eV}}{{2{m_p}eV}}} $

${\lambda _p} = \sqrt 8 {\lambda _\alpha }$

Thus, the de-Broglie wavelength of the proton is $\sqrt 8 $ times of alpha $\left( \alpha  \right)$particle.

15. (i) In the explanation of the photoelectric effect, we assume one photon of frequency\[\;{{\nu }}\] collides with an electron and transfers its energy. This leads to the equation for the maximum energy \[{{{E}}_{{{max}}}}\]of the emitted electron as ${E_{max}} = hv - {\phi _ \circ }$ 

 where ${\phi _0}$ is the work function of the metal. If an electron absorbs 2 photons (each of frequency \[{{\nu }}\]) what will be the maximum energy for the emitted electron? 

(ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?

Ans:   (i) Here, $2$ photons transfer its energy to one electron as

$\therefore {E_e} = {E_p}$

$h{v_e} = 2hv$

$\therefore {v_e} = 2v$

Maximum energy of emitted electron is

${E_{\max }} = h{v_e} - {\phi _0}$

$ = h(2v) - {\phi _0}$

$ = 2hv - {\phi _0}$

(ii) the possibility of absorbing $2$photons by electrons is very low due to their mass difference. So, the probability of such emission of electrons is negligible.

16. There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength. 

Ans:  As we know that as the wavelength of a photon increases, its frequency decreases or energy increases.

Case I: Photons which are of shorter wavelength i.e., those which are of larger energy will effuse the photons of smaller energy. Some energy is utilized against work function. So, it is attainable by law of conservation of energy.

Case II: Photons which are of longer wavelength invariably emit photons of shorter wavelength and photons of smaller energy can never effuse photons of larger energy. Some part of energy $\left( {hv} \right)$ is utilized in the work function $\left( \phi  \right)$ of metal. Or it will repudiate the law of conservation of energy (universal law). So, it is unendurable in stable materials.

17. Do all the electrons that absorb a photon come out as photo-electrons?

Ans:   We can observe that most of the electrons in photoelectric effect pounded by photons are dissipated into the metal by absorbing photons. Thus, some of the energy of photons is absorbed. Generally, one photon cannot emit one electron. Some energy of the photons is consumed against the work function of metal. Thus, some of all electrons which absorb a photon come out from the metal surface.

18. There are two sources of light, each emitting with a power of \[{{100}}{\text{ }}{{W}}\]. One emits X-rays of wavelength \[{{1}}\;{{nm}}\] and the other visible light at \[{{500}}{\text{ }}{{nm}}\]. Find the ratio of the number of photons of X-rays to the photons of visible light of the given wavelength?

Ans:   Let the energies ${E_x}$ and ${E_v}$ given by one photon in X-rays and visible rays then,

${E_x} = h{v_x}$

And ${E_v} = h{v_v}$

Let the number of photons ${n_x}$ and ${n_v}$ X-rays and visible light to give equal energies as both sources (X-rays and visible) emitting same power $100\,W$each so 

${n_x}h{v_x} = {n_v}h{v_v}$

$\dfrac{{{n_x}}}{{{n_v}}} = \dfrac{{{v_v}}}{{{v_x}}}$

$ \Rightarrow \dfrac{{\dfrac{1}{{{\lambda _v}}}}}{{\dfrac{1}{{{\lambda _x}}}}} = \dfrac{{{\lambda _x}}}{{{\lambda _v}}}$$\left( {\because v = \dfrac{c}{\lambda }or\;v = \dfrac{1}{\lambda }} \right)$

$\dfrac{{{n_x}}}{{{n_v}}} = \dfrac{{1\,nm}}{{500\;nm}}$$\left( {\because {\lambda _x} = 1\,nm\;and\,{\lambda _v} = 500\,nm} \right)$

${n_x}:{n_v} = 1:500$

SHORT ANSWER TYPE

19. Consider \[{{Fig}}.{{11}}.{{1}}\]for photo-emission. How would you reconcile with momentum conservation? Note light (photons) have momentum in a different direction than the emitted electrons.

Ans:  The transfer of momentum of photons to atoms of metal when photons pound to the metal surface by reducing its own speed up to zero. Photons convey this momentum to the nucleus and electrons of the metal. The direction of emission of excited electrons is roughly contrary to the direction of photons. The total momentum transferred by the photons will be equivalent to the momentum of all electrons and nucleus.

20. Consider a metal exposed to light of wavelength\[{{600}}{\text{ }}{{nm}}\]. The maximum energy of the electron doubles when light of wavelength \[{{400}}{\text{ }}{{nm}}\]is used. Find the work function in\[{{eV}}\].

Ans:   let ${K_1}$ and ${K_2}$ are the maximum energies of emitted electrons when $600\;nm$and $400\,nm$visible light is used according to the question

${K_2} = 2{K_1}$

${K_{\max }} = hv - \phi $

$ = \dfrac{{hc}}{\lambda } - \phi $

${k_1} = \dfrac{{hc}}{{{\lambda _1}}} - \phi $

${k_2} = \dfrac{{hc}}{{{\lambda _2}}} - \phi $

$ = 2{K_1}$

$\dfrac{{hc}}{{{\lambda _2}}} - \phi  = 2\left[ {\dfrac{{hc}}{{{\lambda _1}}} - \phi } \right]$

$ = \dfrac{{2hc}}{{{\lambda _1}}} - 2\phi $

$\phi  = hc\left[ {\dfrac{2}{{{\lambda _1}}} - \dfrac{1}{{{\lambda _2}}}} \right]$$\left( {\because hc = 1240\,ev\,nm} \right)$

$\because \phi  = 1240\left[ {\dfrac{2}{{600}} - \dfrac{1}{{400}}} \right]eV$

$ = \dfrac{{1240}}{{200}}\left[ {\dfrac{2}{3} - \dfrac{1}{2}} \right]$

$ = 6.2\dfrac{{\left( {4 - 3} \right)}}{6}$

The work function $\phi  = \dfrac{{6.2}}{6}$

$ = 1.03\,eV$

21. Assuming an electron is confined to a \[{{1nm}}\] wide region, find the uncertainty in momentum using Heisenberg Uncertainty principle $\left( {\Delta x \times \Delta p \cong h} \right)$. You can assume the uncertainty in position \[{{\Delta x}}\]as\[\;{{1nm}}\]. Assuming\[p \approx \Delta p\], find the energy of the electron in electron volts.

Ans:   As we know that electron revolves in circular path

Thus, $\Delta r = 1\;nm$

$ = {10^{ - 9}}m$

$\Delta x \times \Delta p \cong h$$\left[ {Given} \right]$

$\Delta p = \dfrac{h}{{\Delta x}}$

$ = \dfrac{{6.62 \times {{10}^{ - 34}}}}{{2\pi \Delta r}}JS$

$\therefore \Delta p = \dfrac{{6.62 \times {{10}^{ - 34}}}}{{2 \times 3.14 \times {{10}^{ - 9}}}}kg\,m\,{s^{ - 1}}$

$ \Rightarrow \Delta p = \dfrac{{331}}{{314}} \times {10^{ - 25}}$

$\therefore E = \dfrac{1}{2}m{v^2}$

$ = \dfrac{{{m^2}{v^2}}}{{2m}}$

$ = \dfrac{{\Delta {p^2}}}{{2m}}$

$ = \dfrac{{331 \times 331 \times {{10}^{ - 50 + 31}}}}{{314 \times 314 \times 18.2}}J$

$ = \dfrac{{331 \times 331 \times {{10}^{ - 19}} \times 1.6 \times {{10}^{ - 19}}}}{{314 \times 314 \times 18.2}}e$

$ = \dfrac{{331 \times 331 \times 16}}{{314 \times 314 \times 182}}$

$ = 3.8 \times {10^{ - 2}}eV$

22. Two monochromatic beams \[{{A}}\]and \[{{B}}\]of equal intensity\[{{I}}\], hit a screen. The number of photons hitting the screen by beam is twice that by beam\[{{B}}\]. Then what inference can you make about their frequencies?

Ans:  Given,

${I_A} = {I_B}$

${n_A}h{v_A} = {n_B}h{v_B}$

$\because {n_A} = 2{n_B}$

$\therefore 2{n_B}{v_A} = {n_B}{v_B}$

$ \Rightarrow 2{v_A} = {v_B}$

Thus, the frequency of the source $B$is twice of the frequency of source $A$.

23. Two particles A and B of de Broglie wavelengths \[{{{\lambda }}_{{1}}}\]and \[{{{\lambda }}_{{2}}}\]combine to form a particle\[{{C}}\]. The process conserves momentum. Find the de-Broglie wavelength of the particle \[{{C}}\]. (The motion is one dimensional).

Ans:  According to the de-Broglie wavelengths

Or $p = \dfrac{h}{\lambda }$

$ \Rightarrow {p_1} = \dfrac{h}{{{\lambda _1}}}$, ${p_2} = \dfrac{h}{{{\lambda _2}}}$and ${p_3} = \dfrac{h}{{{\lambda _3}}}$

${p_1} + {p_2} = {p_3}$

$\dfrac{h}{{{\lambda _1}}} + \dfrac{h}{{{\lambda _2}}} = \dfrac{h}{{{\lambda _3}}}$(${\lambda _3} = the\;wavelength\;of\,particle\;C$)

$\dfrac{1}{{{\lambda _1}}} + \dfrac{1}{{{\lambda _2}}} = \dfrac{1}{{{\lambda _3}}}$

$\dfrac{{{\lambda _2} + {\lambda _1}}}{{{\lambda _1}{\lambda _2}}} = \dfrac{1}{{{\lambda _3}}}$

Case I: When ${p_1}$and ${p_2}$ are the positive then

${\lambda _3} = \dfrac{{{\lambda _2}{\lambda _1}}}{{{\lambda _1} + {\lambda _2}}}$ 

Case II: When both are negative then

${\lambda _3} = \left| {\dfrac{{ - {\lambda _2}{\lambda _1}}}{{{\lambda _1} + {\lambda _2}}}} \right| = \dfrac{{{\lambda _2}{\lambda _1}}}{{{\lambda _1} + {\lambda _2}}}$

Case III: ${p_A} > 0$, ${p_B} > 0$

$\dfrac{h}{{{\lambda _3}}} = \dfrac{h}{{{\lambda _1}}} - \dfrac{h}{{{\lambda _2}}}$

Or $\dfrac{1}{{{\lambda _3}}} = \dfrac{{{\lambda _2} - {\lambda _1}}}{{{\lambda _1}{\lambda _2}}}$

${\lambda _3} = \dfrac{{{\lambda _1}{\lambda _2}}}{{{\lambda _2} - {\lambda _1}}}$

Case IV: ${p_A} < 0$, ${p_B} > 0$

$\dfrac{h}{{{\lambda _3}}} = \dfrac{h}{{{\lambda _2}}} - \dfrac{h}{{{\lambda _1}}}$

$ \Rightarrow \dfrac{h}{{{\lambda _3}}} = \dfrac{{\left( {{\lambda _1} - {\lambda _2}} \right)h}}{{{\lambda _1}{\lambda _2}}}$

$\therefore {\lambda _3} = \dfrac{{{\lambda _1}{\lambda _2}}}{{{\lambda _1} - {\lambda _2}}}$

24. A neutron beam of energy \[{{E}}\] scatters from atoms on a surface with a spacing\[{{d}}{\text{ }} = {\text{ }}{{0}}.{{1}}\;{{nm}}\]. The first maximum of intensity in the reflected beam occurs at\[{{\theta }}{\text{ }} = {\text{ }}{{30}}^\circ \]. What is the kinetic energy \[{{E}}\]of the beam in\[{{eV}}\]?

Ans:  According to the Bragg’s law of diffraction, condition for $nth$ maxima is

$2d\sin \theta  = n\lambda $

Where,$n = 1$,

So, $\lambda  = 2d\sin \theta $

Where $\theta  = {30^ \circ }$(Given)

$ = 2 \times 0.1 \times {10^{ - 9}}\sin {30^ \circ }\left( {\because d = 0.1\;nm} \right)$

$ = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{{{10}^{ - 10}}}}$

$ = 6.6 \times {10^{ - 24}}kg\,m\;{s^{ - 1}}$

$E = \dfrac{1}{2}m{v^2}$

\[E = \dfrac{{6.6 \times 6.6 \times {{10}^{ - 48}}}}{{2 \times 1.6 \times {{10}^{ - 27}}}}J\]

$ = \dfrac{{66 \times 66 \times {{10}^{ - 48 + 46}}}}{{2 \times 16{\kern 1pt} \; \times 16}}$

$E = \dfrac{{33 \times 33}}{{126}} \times {10^{ - 2}}$

$ = 8.5 \times {10^{ - 2}}$

$ = 0.085\;eV$

LONG ANSWER TYPE

25. Consider a thin target \[\left( {{{1}}{{{0}}^{--{{2}}}}{{m}}{\text{ }}{{square}},{\text{ }}{{1}}{{{0}}^{--{{3}}}}{{m}}{\text{ }}{{thickness}}} \right)\]of sodium, which produces a photo-current of \[100\;\mu A\] when a light of intensity \[{\text{100 W }}\;{m^{ - 2}}{\text{ }}\left( {\lambda  = 660\;nm} \right)\]falls on it. Find the probability that a photo-electron is produced when a photon strikes a sodium atom. \[\left[ {{{Take}}{\text{ }}{{density}}{\text{ }}{{of}}{\text{ }}{{Na}}{\text{ }} = {\text{ }}{{0}}.{{97}}{\text{ }}{{kg}}{{{m}}^{ - {{3}}}}{\text{ }}} \right].\]

Ans:  Area of the square sheet ($A$) $ = {10^{ - 2}} \times {10^{ - 2}}$

$ = {10^{ - 4}}\;{m^2}$

Thickness $\left( d \right)$$ = {10^{ - 3}}\;m$

Current $\left( i \right)$$ = 100\;\mu A = {10^{ - 4}}A$

Intensity $\left( I \right)$$ = 100\;W\;{m^{ - 2}}$

Mass of target $\left( m \right)$=$Volume \times density$

$m = area\;of\,sheet \times thickness \times density$

$ \Rightarrow m = \left( {{{10}^{ - 4}} \times {{10}^{ - 3}}} \right) \times 0.97\;kg$

$ \Rightarrow m = 0.97 \times {10^{ - 7}}kg$

$ \Rightarrow m = 0.97 \times {10^{ - 4}}gm$ $\therefore number\,of\;Na\;atoms\,in\,target = \dfrac{{6.023 \times {{10}^{23}}}}{{23}} \times 0.97 \times {10^{ - 4}}$

$ = 0.254 \times {10^{19}}$

Number of $Na$atoms in $Na$target $ = 2.54 \times {10^{18}}atoms$

Total energy falling per second on target $ = nhv$

\[Intensity \times Area = n \times h \times \dfrac{c}{\lambda }\]

$I \times A = \dfrac{{nhc}}{\lambda }$

$n = \dfrac{{IA\lambda }}{{hc}}$

$ = \dfrac{{100 \times {{10}^{ - 4}} \times 660 \times {{10}^{ - 9}}}}{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}$

$ = \dfrac{{1000 \times 660}}{{66 \times 3}} \times {10^{ - 13 - 8 + 34}}$

$ = \dfrac{{10000}}{3} \times {10^{ - 24 + 34}}$

$ = \dfrac{{10}}{3} \times {10^3} \times {10^{13}}$

Number of photons $\left( n \right)$incident per second on $Na$metal $ = 3.3 \times {10^{16}}$

Assuming $P$is the possibility of emission of photo-electrons per atom per photon.

Number of photo-electrons emitted per second$\left( N \right) = P \cdot n$( here $n$ is no. of sodium atoms)

$N = P \times 3.3 \times {10^{16}} \times 2.54 \times {10^{18}}$

 $i = 100\mu A = {10^{ - 4}}A = Ne$

$ \Rightarrow N = \dfrac{i}{e}$

$P = \dfrac{i}{{e \times 3.3 \times {{10}^{16}} \times 2.54 \times {{10}^{18}}}}$

$P = \dfrac{{{{10}^{ - 4}}}}{{1.6 \times {{10}^{ - 19}} \times 3.3 \times {{10}^{16}} \times 2.54 \times {{10}^{18}}}}$

$ = \dfrac{{{{10}^{ - 4 - 34 + 19}}}}{{13.4}}$

$0.075 \times {10^{ - 19}}$

$P = 7.5 \times {10^{21}}$

26. Consider an electron in front of a metallic surface at a distance d (treated as an infinite plane surface). Assume the force of

attraction by the plate is given as $d = 0.1nm$

Ans: According to the question, $F = \dfrac{1}{4}.\dfrac{{{q^2}}}{{4\pi {\varepsilon _0}{d^2}}}$

Assuming that at any instant electron is at distance x from the metal

surface. Attractional force between metal surface and electron is F.

\[F=\frac{1}{4} \times \frac{q^2}{4\pi\varepsilon _0 x^2}\]

Work done by external in taking

the electron from distance d to infinity is 

$W.D. = \int\limits_d^\infty  {F.dx} $

$ \Rightarrow W.D. = \int\limits_d^\infty  {\dfrac{1}{4} \times \dfrac{{{q^2}}}{{4\pi {\varepsilon _0}{x^2}}}dx} $

\[\Rightarrow W.D. = \frac{1}{4}\times \frac{q^2}{4\pi \varepsilon _0}\int\limits_d^\infty x^{-2}dx\]

\[ \Rightarrow W.D. = \dfrac{1}{4} \times \dfrac{{{q^2}}}{{4\pi {\varepsilon _0}}}\left[ { - \dfrac{1}{x}} \right]_d^\infty \]

\[ \Rightarrow W.D. = \dfrac{1}{4} \times \dfrac{{{q^2}}}{{4\pi {\varepsilon _0}}}\left[ { - \dfrac{1}{\infty } + \dfrac{1}{d}} \right]\]

\[ \Rightarrow W.D. = \dfrac{{k{q^2}}}{{4d}}\]

Where $d = 0.1nm = {10^{ - 10}}m$

Work done here is positive

So,  \[W.D.= \frac{9 \times 10^9 \times \left(1.6 \times 10^{-19}\right)^2}{4 \times 10^{-10}}J\]

\[ \Rightarrow W.D. = 5.76 \times {10^{ - 19}}J\]

Converting to eV, we get,

\[ \Rightarrow W.D. = \dfrac{{5.76 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}}J\]

\[ \Rightarrow W.D. = \dfrac{{5.76}}{{1.6}}eV\]

\[ \Rightarrow W.D. = 3.6\,eV\]

27. A student performs an experiment on the photoelectric effect, using two materials \[{{A}}\]and\[{{B}}\]. A plot of \[{{\text{V}}_{_{{{stop}}}}}{\text{ vs}}\;{\text{v}}\]is given in Fig. 11.2.  

(i) Which material \[{{A}}\] or \[{{B}}\] has a higher work function? 

(ii) Given the electric charge of an electron\[ = {\text{ }}{{1}}.{{6}}{\text{ }} \times {\text{ }}{{1}}{{{0}}^{--{{19}}}}{\text{ }}{{C}}\], find the value of \[{{h}}\]obtained from the experiment for both \[{{A}}\]and \[{{B}}\]. Comment on whether it is consistent with Einstein’s theory:

Ans:   (i) As we know

${\phi _ \circ } = h{v_ \circ }$

Where ${v_ \circ } = $ Threshold frequency 

$h = $ Planck’s constant 

For metal \[A{\text{ }}{v_{ \circ A}} = {\text{ }}5 \times {10^4}{\text{ }}Hz\]

For metal \[B{\text{ }}{v_{ \circ B}} = 10 \times {10^4}{\text{ }}Hz\]

$\dfrac{{{\phi _{ \circ A}}}}{{{\phi _{ \circ B}}}} = \dfrac{{h{v_{ \circ A}}}}{{h{v_{ \circ B}}}}$

$ = \dfrac{{5 \times {{10}^{14}}}}{{10 \times {{10}^{14}}}}$

$\therefore {\phi _{ \circ B}} = 2{\phi _{ \circ A}}$

Thus, the work function of material\[\;B\] is twice of material\[A\]. 

(ii) By the differentiation of potential 

$V = \dfrac{W}{Q}$

If \[W = E\]and \[Q = e\](charge on an electron) 

$V = \dfrac{E}{e}$

Or\[\;E = eV\] 

\[hv{\text{ }} = {\text{ }}eV{\text{ }}\left( {\because {\text{ }}E = {\text{ }}hv} \right)\]

Differentiating both sides, we get 

\[h.dv = {\text{ }}edV\]

$h = e \cdot \dfrac{{dV}}{{dv}}$

For metal A, 

$h = \dfrac{{1.6 \times {{10}^{ - 19}}\left[ {2 - 0} \right]}}{{\left( {10 - 5} \right) \times {{10}^{14}}}}$

$ = \dfrac{{3.2}}{5} \times {10^{ - 19 + 14}}$

\[ = 0.64 \times {10^{ - 33}}JS\]

\[ = {\text{ }}6.4 \times {10^{ - 33}}{\text{ }}JS\]

\[h = 6 \times {10^{ - 34}}{\text{ }}JS{\text{ }} \ldots  \ldots  \ldots \left( I \right)\]

For metal B, 

$h = \dfrac{{e \times \left( {2.5 - 0} \right)}}{{\left( {10 - 5} \right) \times {{10}^{14}}}}$

$ = \dfrac{{1.6 \times {{10}^{ - 19}} \times 2.5 \times {{10}^{ - 14}}}}{5}$

$ = \dfrac{{4.00}}{5} \times {10^{ - 33}}$

$ = 0.8 \times {10^{ - 33}}$

\[h{\text{ }} = 8 \times {10^{ - 34{\text{ }}}}JS{\text{ }} \ldots .{\text{ }}\left( {II} \right)\]

The Planck’s constant \[\left( h \right)\]’s value for both experimental graphs are not equal, so the experiment is not consistent with Einstein’s theory. 

But because of experimental limitations, values are very close to\[6.6 \times {10^{ - 34}}{\text{ }}JS\], so can be considered consistent with Einstein theory. 

28. A particle \[{{A}}\]with a mass \[{{mA}}\]is moving with a velocity\[\;{{v}}\] and hits a particle\[\;{{B}}\] (mass\[{{mB}}\]) at rest (one dimensional motion). Find the change in the de-Broglie wavelength of the particle\[{{A}}\]. Treat the collision as elastic.

Ans:  Because the collision is elastic, the law of conservation of momentum and kinetic energy are followed.

${m_A}v + {m_B}\left( 0 \right) = {m_A}{v_1} + {m_B}{v_2}$

${m_A}\left( {v - {v_1}} \right) = {m_B}{v_2}.....\left( I \right)$

And $\dfrac{1}{2}{m_A}{v^2} + \dfrac{1}{2}{m_B}{\left( 0 \right)^2} = \dfrac{1}{2}{m_A}v_1^2 + \dfrac{1}{2}{m_B}v_2^2$

${m_A}\left( {{v^2} - v_1^2} \right) = {m_B}v_2^2$

${m_A}\left( {v - v_1^{}} \right)\left( {v + v_1^{}} \right) = {m_B}v_2^2.....(II)$

On dividing (II) by(I) we get,

$v + {v_1} = {v_2}......(III)$

$v = {v_2} - {v_1}$

Putting (III) in (I)

${m_A}v - {m_A}{v_1} = {m_B}\left( {v + {v_1}} \right)$

${m_A}v - {m_A}{v_1} = {m_B}v + {m_B}{v_1}$

${m_A}v - {m_B}v = {m_A}{v_1} + {m_B}{v_1}$

$\left( {{m_A} - {m_B}} \right)v = \left( {{m_A} + {m_B}} \right){v_1}$

${v_1} = \dfrac{{\left( {{m_A} - {m_B}} \right)}}{{\left( {{m_A} + {m_B}} \right)}}v$

${v_2} = v + \dfrac{{\left( {{m_A} - {m_B}} \right)}}{{\left( {{m_A} + {m_B}} \right)}}v$

$ = v\left[ {1 + \dfrac{{\left( {{m_A} - {m_B}} \right)}}{{\left( {{m_A} + {m_B}} \right)}}} \right]$

$ = v\left[ {\dfrac{{\left( {{m_A} + {m_B} + {m_A} - {m_B}} \right)}}{{\left( {{m_A} + {m_B}} \right)}}} \right]$

$ = \dfrac{{2{m_A}v}}{{\left( {{m_A} + {m_B}} \right)}}$

${\lambda _{A\;initial}} = \dfrac{h}{{{m_A}v}}$

${\lambda _{A\;final}} = \dfrac{h}{{{m_A}v}}$

${\lambda _{A\;final}} = \dfrac{{h\left( {{m_A} + {m_B}} \right)}}{{{m_A}\left( {{m_A} - {m_B}} \right)v}}$

$\Delta \lambda  = {\lambda _{A\;final}} - {\lambda _{A\;initial}}$

$ = \dfrac{h}{{{m_A}v}}\left[ {\dfrac{{\left( {{m_A} + {m_B}} \right)}}{{\left( {{m_A} - {m_B}} \right)}} - 1} \right]$

$ = \dfrac{h}{{{m_A}v}}\left[ {\dfrac{{\left( {{m_A} + {m_B}} \right) - \left( {{m_A} - {m_B}} \right)}}{{\left( {{m_A} - {m_B}} \right)}}} \right]$

Thus, the change in the de-Broglie wavelength $\Delta \lambda  = \dfrac{{2{m_B}h}}{{{m_A}\left( {{m_A} - {m_B}} \right)v}}$

29. Consider a \[{{20}}{\text{ }}{{W}}\]bulb emitting light of wavelength \[{{5000}}\mathop {\rm A}\limits^ \circ  \] and shining on a metal surface kept at a distance\[{{2m}}\]. Assume that the metal surface has work function of \[{{2}}{\text{ }}{{eV}}\]and that each atom on the metal surface can be treated as a circular disk of radius\[{{1}}.{{5}}\mathop {\rm A}\limits^ \circ  \].

(i) Estimate no. of photons emitted by the bulb per second. [Assume no other losses] 

(ii) Will there be photoelectric emission? 

(iii) How much time would be required by the atomic disk to receive energy equal to work function (\[{{2}}{\text{ }}{{eV}}\])? 

(iv) How many photons would an atomic disk receive within the time duration calculated in (iii) above? 

(v) Can you explain how the photoelectric effect was observed instantaneously?

Ans:   Given, $P = 20\;W$, $\lambda  = 5000\;\mathop {\rm A}\limits^ \circ   = 5000 \times {10^{ - 10}}\;m$, $r = 1.5\;\mathop {\rm A}\limits^ \circ   = 1.5 \times {10^{ - 10}}\;m$(atomic radius) and $\phi  = 2\;eV$

Assuming the number of photons emitted by bulb per second is ${n_1}$ then Power $P$ is 

$P = {n_1}hv$

Or $P = {n_1}\dfrac{c}{\lambda }$

$\therefore {n_1} = \dfrac{{P\lambda }}{{hc}}$

$ = \dfrac{{20 \times 5000 \times {{10}^{ - 10}}}}{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}$

${n_1} = \dfrac{{100000 \times {{10}^{ - 10 - 8 + 34}}}}{{19.86}} \approx \dfrac{{100000}}{{20}} \times {10^{ - 18 + 34}}$

${n_1} = 5 \times {10^{16 + 3}}$

$ = 5 \times {10^{19}}$(no. of photons per second)

Number of photons effused by bulb per second

${n_1} = 5 \times {10^{19}}$

(ii) incident photon’s energy

$ = \dfrac{{hc}}{\lambda }$

$ = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{5000 \times {{10}^{ - 10}}}}J$

$ = \dfrac{{19.86 \times {{10}^{ - 34 + 10 + 8}}}}{{5000 \times 1.6 \times {{10}^{ - 19}}}}eV$

$ = 2.48\,eV$

Energy of photon $ = hv$

$ = \dfrac{{\left( {6.62 \times {{10}^{ - 34}}} \right)\left( {3 \times {{10}^8}} \right)}}{{5000 \times {{10}^{ - 10}}}}J$

$ \Rightarrow \dfrac{{6.62 \times 3 \times {{10}^{ - 24 + 8}}}}{{5000 \times 1.6 \times {{10}^{ - 19}}}}eV = \dfrac{{19.86 \times {{10}^{ - 16 + 19}}}}{{8000}}$

$Energy\;of\;photon = \dfrac{{20 \times {{10}^3}}}{{8 \times {{10}^3}}}$

$ = \dfrac{5}{2}eV$

The photoelectric emission happened because the energy of an incident photon is comparably more than $2\;eV$i.e., metal surface’s work function.

(iii) Assuming $\Delta T$time depleted in getting the energy \[\phi \] (work function of metal) energy sustained by atomic disk in

At time $\Delta t$,

$E = P \times A \cdot \Delta t$

Energy transmitted by bulb in full solid angle $4\pi {d^2}$to atoms $ = 4\pi {d^2}\phi $

$\therefore p \times \pi {r^2}\Delta t = 4\pi {d^2}\phi $

$\Delta t = \dfrac{{4{d^2}\phi }}{{{{\Pr }^2}}}\dfrac{{4 \times 2 \times 2 \times 2 \times 1.6 \times {{10}^{ - 19}}}}{{20 \times 1.5 \times 1.5 \times {{10}^{ - 10}} \times {{10}^{ - 10}}}}\sec $

$\Delta t = \dfrac{{12.8 \times {{10}^{ - 19 + 20}}}}{{5 \times 2.25}}$

$ = \dfrac{{128}}{{12.25}}$

$ = 11.4\;\sec $

(iv) Number of photons acquired by one atomic disk in time $\Delta t$

$N = \dfrac{{{n_1}{r^2}\Delta t}}{{4\pi {d^2}}}$

$ = \dfrac{{{n_1}{r^2}\Delta t}}{{4{d^2}}}$

$ = \dfrac{{5 \times {{10}^{19}} \times 1.5 \times 1.5 \times {{10}^{ - 20}} \times 11.4}}{{4 \times 2 \times 2}}$[${n_1}$from part (i) from part (iii)]

$N = \dfrac{{12.25 \times 11.4 \times {{10}^{ - 1}}}}{{8 \times 2 \times 2}} \approx 0.80 \cong 1\;photon\,per\,atom$

$N = 1\,photon\;per\,atom$.

(v) $11.4\;\sec $   is the time of emission of electrons. Thus, photoelectric emission is not an immediate problem. Thus, it takes about $11.4\,\sec $. There is a collision among photon and free electron and nucleus in photoelectric which lasts for very short time interval $\left( {{{10}^{ - 9}}\sec } \right)$therefore we say photoelectric emission is expeditious. 

NCERT Solutions For Class 12 Physics Chapter 11

Chapter 11 – dual nature of radiation and matter.

NCERT Textbooks, e-Content, and Study Materials are Copyright of The National Council of Educational Research and Training . CBSE/ICSE/NIOS & Other Education Board's e-Contents are wholly owned by respective States Governments or the Ministry of Education, India . We do not claim any type of ownership of these materials nor commercially promote them. Please contact us before taking any legal action, we will remove all links related to the copyright owner.

AssignmentsBag.com

Assignments For Class 12 Physics

Solve these Physics Assignment for Class 12 PDF with answers. These Physics Assignment for Class 12 PDF are prepared by our expert teachers on the latest exam pattern of the CBSE Board Exam. We have provided you CBSE Class 12 Physics Assignment Pdf with answers to help students to make their preparation better.

Assignments for Class 12 Physics have been developed for Standard 12 students based on the latest syllabus and textbooks applicable in CBSE, NCERT and KVS schools. Parents and students can download the full collection of class assignments for class 12 Physics from our website as we have provided all topic wise assignments free in PDF format which can be downloaded easily. Students are recommended to do these assignments daily by taking printouts and going through the questions and answers for Grade 12 Physics. You should try to do these test assignments on a daily basis so that you are able to understand the concepts and details of each chapter in your Physics book and get good marks in class 12 exams.

Assignments for Class 12 Physics as per CBSE NCERT pattern

All students studying in Grade 12 Physics should download the assignments provided here and use them for their daily routine practice. This will help them to get better grades in Physics exam for standard 12. We have made sure that all topics given in your textbook for Physics which is suggested in Class 12 have been covered ad we have made assignments and test papers for all topics which your teacher has been teaching in your class. All chapter wise assignments have been made by our teachers after full research of each important topic in the textbooks so that you have enough questions and their solutions to help them practice so that they are able to get full practice and understanding of all important topics. Our teachers at https://www.assignmentsbag.com have made sure that all test papers have been designed as per CBSE, NCERT and KVS syllabus and examination pattern. These question banks have been recommended in various schools and have supported many students to practice and further enhance their scores in school and have also assisted them to appear in other school level tests and examinations. Its easy to take print of thee assignments as all are available in PDF format.

Some advantages of Free Assignments for Class 12 Physics

• Solving Assignments for Physics Class 12 helps to further enhance understanding of the topics given in your text book which will help you to get better marks
• By solving one assignments given in your class by Physics teacher for class 12 will help you to keep in touch with the topic thus reducing dependence on last minute studies
• You will be able to understand the type of questions which are expected in your Physics class test
• You will be able to revise all topics given in the ebook for Class 12 Physics as all questions have been provided in the question banks
• NCERT Class 12 Physics Workbooks will surely help you to make your concepts stronger and better than anyone else in your class.
• Parents will be able to take print out of the assignments and give to their child easily.

All free Printable practice assignments are in PDF single lick download format and have been prepared by Class 12 Physics teachers after full study of all topics which have been given in each chapter so that the students are able to take complete benefit from the worksheets. The Chapter wise question bank and revision assignments can be accessed free and anywhere. Go ahead and click on the links above to download free CBSE Class 12 Physics Assignments PDF.

Free PDF download of Physics Assignment for Class 12 pdf created by master educators from the latest syllabus of CBSE Boards. By practicing these Assignment of physics Class 12 will help you to score more marks in your CBSE Board Examinations. We also give free NCERT Solutions and other study materials for students to make their preparation better.

Students who are searching for better solutions can download the Physics Assignment for Class 12 PDF with solutions to assist you with revising the whole syllabus and score higher marks in your exam.

This Chapter wise assignment for Class 12 physics with solutions shows up with an answer key with step-by-step answers for students to comprehend the problem at each level and ​not retain it.

We hope this Physics Assignment for Class 12 PDF with answers shared with you will help you to score good marks in your exam. We have also other study material like NCERT Solutions, NCERT Book, Exam Question, and simpler paper of Class 12. If You want to score higher in your exam, So practice all this study material and if you have any problem then, write it in the comment box and we will guide you as much as possible.

You can download free assignments for class 12 Physics from https://www.assignmentsbag.com

You can get free PDF downloadable assignments for Grade 12 Physics from our website which has been developed by teachers after doing extensive research in each topic.

On our website we have provided assignments for all subjects in Grade 12, all topic wise test sheets have been provided in a logical manner so that you can scroll through the topics and download the worksheet that you want.

You can easily get question banks, topic wise notes and questions and other useful study material from https://www.assignmentsbag.com without any charge

Yes all test papers for Physics Class 12 are available for free, no charge has been put so that the students can benefit from it. And offcourse all is available for download in PDF format and with a single click you can download all assignments.

https://www.assignmentsbag.com is the best portal to download all assignments for all classes without any charges.

Related Posts

Assignments For Class 6 Hindi

Assignments class 12 chemistry coordination compounds.

Assignments For Class 11 Mathematics Limits And Derivatives

NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2

Ncert solutions for class 12 maths chapter 11 three dimensional geometry ex 11.2.

Free PDF of NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2 prepared by expert Mathematics teacher at Mathongo.com as per CBSE (NCERT) books guidelines. Download our Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks in your exams.

Share with friends:

NCERT Solutions for Class 12 Maths Exercise 11.2

NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2 of 3D Three Dimensional Geometry for CBSE 2024-25 exams. Questions of Ex. 11.2 of 12th Maths are revised as per rationalised syllabus and new textbook issued for academic session 2024-25.

Class 12 Maths Exercise 11.2 in Hindi and English Medium

• Class 12 Maths Exercise 11.2 in English
• Class 12 Maths Exercise 11.2 in Hindi
• Class 12 Maths Chapter 11 Solutions
• Class 12 Maths NCERT Solutions
• Class 12 all Subjects NCERT Solutions

The NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2 updated for session 2024-25 are given here in Hindi and English Medium free to use. Videos based on Hindi Medium and English Medium of Exercise 11.2 are available on the page free to download or use online. 12th Maths exercise 11.2 solutions for CBSE and state boards. All the contents are available in Hindi and English Medium free to use online or download in PDF file format free without any login or password.

A line given in space can extend in two opposite directions, thus it contains two-direction cosines. For a given line in space to have a unique set of direction cosines, we must take the given line as a directed line. These unique direction cosines are denoted by L, M and N. If the given line in space does not pass through the origin, to find its cosine direction, we draw a line through the origin and are parallel to the given line.

Now take one of the lines directed from the origin and find its direction cosine, because the two parallel lines have the same set of direction cosines. Any three numbers that are proportional to the direction corners of a line are called the direction relationships of the line. If l, m, n are directions and a, b, c are the direction relations of a line, then a = λl, b = λm and c = λn, for any nonzero real number λ.

Explain that L1 and L2 are two lines that pass through the origin and the direction relation a1, b1, c1 and a2, b2, c2 respectively. Explain that P is a point on L1 and Q is a point on L2. Consider the directed lines OP and OQ. Let A be the acute angle between OP and OQ. Now remember that the directed line segments OP and OQ are vectors with components A1, B1, C1 and A2, B2, C2 respectively. Therefore, the angle A between them is given by the formula CosA.

What are the main concepts on which class 12 Maths Exercise 11.2 is based?

In class 12 Maths Exercise 11.2, we will learn how to find the equation of a line in vector as well as Cartesian plane, angle between two lines and distance between the two lines.

Which is the most important question in Exercise 11.2 of 12th Maths?

Questions number 12 and 15 are the asked so many times in board exams. So, these are important and tricky as well.

Copyright 2024 by Tiwari Academy | A step towards Free Education

NCERT Solutions

• NCERT Class 12
• NCERT 12 Maths
• Chapter 11: 3 Dimensional Geometry
• Exercise 11.2

NCERT Solution Class 12 Chapter 11 - Three Dimensional Geometry Exercise 11.2

Ncert solutions for class 12 maths chapter 11 exercise 11.2 – free pdf download.

Exercise 11.2 of NCERT Solutions for Class 12 Maths Chapter 11 – Three Dimensional Geometry is based on the following topics:

• Equation of a line through a given point and parallel to a given vector b.
• Equation of a line passing through two given points
• Angle between Two Lines
• Distance between two skew lines
• Distance between parallel lines

The exercise contains problems that are based on the topics mentioned above. Hence, solving this exercise helps the students understand the concepts covered in the chapter thoroughly.

NCERT Solutions for Class 12 Maths Chapter 11 – Three Dimensional Geometry Exercise 11.2

carouselExampleControls112

Previous Next

Access Other Exercises of Class 12 Maths Chapter 11

Exercise 11.1 Solutions 5 Questions

Exercise 11.3 Solutions 14 Questions

Miscellaneous Exercise on Chapter 11 Solutions 23 Questions

Access Answers to NCERT Class 12 Maths Chapter 11 Exercise 11.2

1. Show that the three lines with direction cosines

Let us consider the direction cosines of L 1 , L 2  and L 3  be l 1 , m 1 , n 1 ; l 2 , m 2 , n 2  and l 3 , m 3 , n 3 .

We know that

If l 1 , m 1 , n 1  and l 2 , m 2 , n 2 are the direction cosines of two lines,

And θ is the acute angle between the two lines,

Then cos θ = |l 1 l 2  + m 1 m 2  + n 1 n 2 |

If two lines are perpendicular, then the angle between the two is θ = 90°

For perpendicular lines, | l 1 l 2  + m 1 m 2  + n 1 n 2  | = cos 90° = 0, i.e. | l 1 l 2  + m 1 m 2  + n 1 n 2  | = 0

So, in order to check if the three lines are mutually perpendicular, we compute | l 1 l 2  + m 1 m 2  + n 1 n 2  | for all the pairs of the three lines.

First, let us compute, | l 1 l 2  + m 1 m 2  + n 1 n 2  |

So,  L 1 ⊥ L 2  …… (1)

Let us compute, | l 2 l 3  + m 2 m 3  + n 2 n 3  |

So, L 2 ⊥ L 3  ….. (2)

Let us compute, | l 3 l 1  + m 3 m 1  + n 3 n 1  |

So, L 1 ⊥ L 3  ….. (3)

∴ By (1), (2) and (3), the lines are perpendicular.

Hence, L 1 , L 2  and L 3  are mutually perpendicular.

2. Show that the line through the points (1, –1, 2), (3, 4, –2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

The points (1, –1, 2), (3, 4, –2) and (0, 3, 2), (3, 5, 6).

Let us consider AB be the line joining the points (1, -1, 2) and (3, 4, -2), and CD be the line through the points (0, 3, 2) and (3, 5, 6).

The direction ratios, a 1 , b 1 , c 1 of AB are

(3 – 1), (4 – (-1)), (-2 – 2) = 2, 5, -4.

The direction ratios, a 2 , b 2 , c 2 of CD are

(3 – 0), (5 – 3), (6 – 2) = 3, 2, 4.

Then, AB and CD will be perpendicular to each other, if a 1 a 2 + b 1 b 2 + c 1 c 2 = 0

a 1 a 2 + b 1 b 2 + c 1 c 2 = 2(3) + 5(2) + 4(-4)

= 6 + 10 – 16

∴ AB and CD are perpendicular to each other.

3. Show that the line through points (4, 7, 8), (2, 3, 4) is parallel to the line through points (-1, -2, 1), (1, 2, 5).

The points (4, 7, 8), (2, 3, 4) and (–1, –2, 1), (1, 2, 5).

Let us consider AB be the line joining the points (4, 7, 8), (2, 3, 4), and CD be the line through the points (-1, -2, 1), (1, 2, 5).

(2 -4), (3 -7), (4 -8) = -2, -4, -4.

(1 – (-1)), (2 – (-2)), (5 – 1) = 2, 4, 4.

Then, AB will be parallel to CD, if

So, a 1 /a 2 = -2/2 = -1

b 1 /b 2 = -4/4 = -1

c 1 /c 2 = -4/4 = -1

∴ We can say that,

-1 = -1 = -1

Hence, AB is parallel to CD, where the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (-1, -2, 1), (1, 2, 5).

6. Find the Cartesian equation of the line which passes through the point (-2, 4, -5) and parallel to the line given by

The points (-2, 4, -5)

The Cartesian equation of a line through a point (x 1 , y 1 , z 1 ) and having direction ratios a, b, c is

7. The Cartesian equation of a line is

So when comparing this standard form with the given equation, we get

x 1  = 5, y 1  = -4, z 1 = 6, and

l = 3, m = 7, n = 2

8. Find the vector and the Cartesian equations of the lines that pass through the origin and (5, -2, 3).

9. Find the vector and the Cartesian equations of the line that passes through the points (3, -2, -5), (3, –2, 6).

By (3), we have

12. Find the values of p so that the lines

So, the direction ratios of the lines are

-3, 2p/7, 2 and -3p/7, 1, -5

Now, both the lines are at right angles,

So, a 1 a 2  + b 1 b 2  + c 1 c 2  = 0

(-3) (-3p/7) + (2p/7) (1) + 2 (-5) = 0

9p/7 + 2p/7 – 10 = 0

(9p+2p)/7 = 10

∴ The value of p is 70/11.

13. Show that the lines

The equations of the given lines are

Two lines with direction ratios are given as

a 1 a 2  + b 1 b 2  + c 1 c 2  = 0

So, the direction ratios of the given lines are 7, -5, 1 and 1, 2, 3

i.e., a 1  = 7, b 1  = -5, c 1 = 1, and

a 2  = 1, b 2  = 2, c 2  = 3

Now, considering

a 1 a 2  + b 1 b 2  + c 1 c 2  = 7 × 1 + (-5) × 2 + 1 × 3

= 7 -10 + 3

∴ The two lines are perpendicular to each other.

Let us rationalize the fraction by multiplying the numerator and denominator by √2, and we get

∴ The shortest distance is 3 √ 2/2.

∴ The shortest distance is 2 √ 29.

Here, by comparing the equations, we get

∴ The shortest distance is 3 √ 19.

17. Find the shortest distance between the lines whose vector equations are

∴ The shortest distance is 8 √ 29.

Also, explore –

NCERT Solutions for Class 12 Maths

NCERT Solutions for Class 12

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

• Share Share

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.