Physics NCERT Exemplar Solutions Class 12th Chapter Eleven Dual Nature of Radiation and Matter

Explanation- given d= 0.1nm, $\theta$ =30⁰ and n=1

According to bragg’s law

2dsin $\theta$ = n $\lambda$

2 $\times 0.1\times sin30=\lambda$

$\lambda =0.1nm={10}^{-10}$

$\lambda =\frac{h}{mv}=\frac{h}{p}$

P=h/ $\lambda$  = $\frac{6.62\times {10}^{-34}}{{10}^{-10}}$ = 6.62 $\times {10}^{-24}$ kgm/s

k.E= 1/2mv²= $\frac{1}{2}\frac{m^2{v}^2}{m}$ = $\frac{p^2}{2m}$ = 0.21eV

4. Consider a thin target (10^–2m square, 10^–3m thickness) of sodium, which produces a photocurrent of 100μA when a light of intensity 100W/m²(λ = 660nm) falls on it. Find the probability that a photoelectron is produced when a photons strikes a sodium atom. [Take density of Na = 0.97 kg/m³]

Explanation- A= 10^-4m²

So d= 10^-3 and i= 100^-4A

I= 100W/m²

$\mathrm{\lambda }=600\mathrm{n}\mathrm{m}=600\times {10}^{-9}$

${\mathrm{\rho }}_{\mathrm{N}\mathrm{a}}=0.97\mathrm{k}\mathrm{g}/{\mathrm{m}}^3$

$\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}=\mathrm{A}\times \mathrm{d}$ 10^-4(10^-3)=10^-7m³

$\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{}23\mathrm{k}\mathrm{g}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{i}\mathrm{d}\mathrm{u}\mathrm{m}$

So volume Na atoms=23/0.97m³

Volume occupied by one Na atom= $\frac{23}{0.97\times 6\times {10}^{26}}=3.95\times {10}^{-26}{\mathrm{m}}^3$

Number of Na atoms in target $\frac{{10}^{-7}}{3.95\times {10}^{-26}}=2.53\times {10}^{18}$

So energy falling per sec= $\frac{\mathrm{n}\mathrm{h}\mathrm{c}}{\mathrm{\lambda }}=\mathrm{I}\mathrm{A}$

So n= $\frac{\mathrm{I}\mathrm{A}\mathrm{\lambda }}{\mathrm{h}\mathrm{c}}$ = $\frac{100\times {10}^{-4}\times 660\times {10}^{-9}}{6.62\times {10}^{-34}\times 3\times {10}^8}=3.3\times {10}^{16}$

N=P $\times \mathrm{n}\times \mathrm{N}\mathrm{a}=\mathrm{P}\times 3.3\times {10}^{16}\times 2.53\times {10}^{18}$

I = 100 $\times {10}^{-6}={10}^{-4}$ A

I=Ne= $\mathrm{P}\times 3.3\times {10}^{16}\times 2.53\times {10}^{18}$ ( ${10}^{-4}$ A)

P= 7.48 $\times {10}^{-21}$ it is less than 1.

5. Consider an electron in front of metallic surface of a distance d. Assume the force of attraction by the plate is given as $\frac{1}{4}\frac{{\mathit{q}}^2}{4\mathit{\pi }{\mathit{\epsilon }}_0{\mathit{d}}^2}$ . Calculate work in taking the to an infinite distance from the plate .Taking d=0.1 nm. find the work done in electron volts?

Explanation- F_x= $\frac{1}{4}\frac{q^2}{4\pi {\epsilon }_0{x}^2}$

W= ${\int }_d^{\infty }fdx$ = ${\int }_d^{\infty }\frac{q^{2}dx}{4\times 4\pi {\epsilon }_0}\frac{1}{x^2}$

= $\frac{q^2}{4\times 4\pi {\epsilon }_0}\frac{1}{d}$

= $\frac{{(1.6\times {10}^{-19})}^2\times 9\times {10}^9}{4\times {10}^{-10}}$ J= 3.6eV 

6. A student performs an experiment on photoelectric effect, using two materials A and B. A plot of V_stopversus v is given in figure.

Which has higher work function?

Explanation -Given threshold frequency of A is given by v_0A= 5 $\times {10}^{14}$ hz

V_OB= 10 $\times$ 10¹⁴hz

$\varnothing =h{v}_0$

$\frac{{\varnothing }_{OA}}{{\varnothing }_{OB}}=\frac{5\times {10}^{14}}{10\times {10}^{14}}1$

${\varnothing }_{OA}$ < ${\varnothing }_{OB}$

$\left(ii\right)$  for metal A slope=h/e= $\frac{2}{(10-5){10}^{14}}$

$h=\frac{2e}{5\times {10}^{14}}=\frac{2\times 1.6\times {10}^{-19}}{5\times {10}^{14}}$ = 6.4 $\times {10}^{-34}$ js

$\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{B}$ slope=h/e= $\frac{2.5}{(15-10){10}^{14}}$ = 8 $\times {10}^{-34}$ js

Explanation- according to law of conservation of momentum

S0 m_Av+m_b0=m_Av₁+m_Bv₂

So m_A(v-v₁)= m_Bv₂

according to law of conservation of kinetic energy

1/2m_Av²=1/2m_Av₁²+1/2m_Bv₂²

So m_A(v²-v₁²)= m_Bv₂²

From above eqn we can say that v+v₁=v₂ or v=v₂-v₁

So v₁= $\frac{m_A{-m}_B}{m_A{+m}_B}$ v  and v₂= $\frac{2m_A}{m_A{+m}_B}$ v

_{$\lambda$ initial}=h/m_Av

_{$\lambda$ final=}h/m_Av₁= $\frac{h\left(m_A{+m}_B\right)}{m_{a\left(m_A{-m}_B\right)v}}$

$d\lambda$ = $\lambda$ _final- $\lambda$ _initial= $\frac{h}{m_{A}v}\{\frac{m_A{+m}_B}{m_A{-m}_B}-1\}$

8. Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance 2 m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å.
(i) Estimate number 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 eV)?
(iv) How many photons would atomic disk receive within time duration calculated in (iii) above?
(v) Can you explain how photoelectric effect was observed instantaneously?

Explanation-number of photon emitted per second n= $\frac{p}{\frac{hc}{\lambda }}=\frac{p\lambda }{hc}=\frac{20\times 5000\times {10}^{-10}}{6.62\times {10}^{-34}\times 3\times {10}^8}$

$=5\times {10}^{19}{s}^{-1}$

(ii) E=hc $/\lambda$ = $\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{5000\times {10}^{-10}\times 1.6\times {10}^{-19}}=2.48eV$  this enegy is greater than 2 so emission is possible

(iii) work function $\varnothing$ = $\frac{p}{4\pi d^2}\times \pi r^2∆t$ = ${\varnothing }_o$

$∆t$ = $\frac{4\varnothing d^2}{p{r}^2}$ = $\frac{4\times 2\times 16\times 1.6\times {10}^{-19}\times 2^2}{20\times {(1.5\times {10}^{-10})}^{-2}}=28.4s$

(iv) N= $\frac{n\pi r^2}{4\pi d^2}\times ∆t$

 = $\frac{5\times {10}^{19}\times {(1.5\times {10}^{-10})}^2\times 28.4}{4\times ({2)}^2}$ =2

(v) as the time of emission is 11.04s so photoelectric is not spontaneous.

1. A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to  (a) H  (b) H^1/2 (c) H⁰   (d) H^-1/2

Explanation- velocity of a freely falling body is v= $\sqrt[]{2gh}$

And $\lambda =\frac{h}{mv}=\frac{h}{m\sqrt[]{2gh}}$

$\lambda =h$ ^-1

Answer –(b)

Explanation- E=hc/ $\lambda$

And l = hc/E= 1240/10⁶= 1.24 $\times$ 10^-3nm

3. Consider a beam of electrons (each electron with energy E₀) 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, E₀
(c) electrons can be emitted with any energy, with a maximum of E₀– ɸ (ɸ  is the work function)
(d) electron can be omitted with energy ,with a maximum of E₀

answer-(d)

explanation- When a beam of electrons of energy E₀ is incident on a metal surface kept in an

evacuated chamber electrons can be emitted with maximum energy E₀ (due to elastic

collision) and with any energy less than E₀, when part of incident energy of electron is

used in liberating the electrons from the surface of metal.

4. Consider the figure given below. Suppose the voltage applied to A is increased. The diffracted beam will have the maximum at a value of θ 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

Answer-(c)

Explanation- In Davisson-Germer experiment, the de-Broglie wavelength associated with electron is

Å ...(i)where V is the applied voltage.

If there is a maxima of the diffracted electrons at an angle θ, then

2dsinθ = λ …(ii)

From Eq.(i),we note that if V is inversely proportional to the wavelength λ.

i.e., V will increase with the decrease in the λ.

From Eq.(ii),we note that wavelength λ is directly proportional to sinθ and hence θ.

So, with the decrease in λ, θ will also decrease.

Thus, when the voltage applied to A is increased. The diffracted beam will have the

Maximum at a value of θ that will be less than the earlier value.

5. A proton, a neutron, an electron and an a-particle have same energy. Then, their de-Broglie wavelengths compare as (a) λ_p= λ_n > λ_e > λ_α     (b) λ_α < λ_p = λ_n > λ_e
(c) λ_e< λ_p = λn> λ_α     (d) λ_e = λ_p = λn = λ_α

answer-(b)

Explanation-as we know that $\lambda =\frac{h}{\sqrt[]{2mk}}$

so $\lambda$ inversely proportional to $\sqrt[]{m}$

6. An electron is moving with an initial velocity v = v₀i and is in a magnetic field B = B₀ Then, its de-Broglie wavelength
(a) remains constant
(b) increases with time
(c) decreases with time
(d) increases and decreases periodically

Answer-(a)

Explanation- v=v₀i and B =B₀j

Force on moving electron F= e(v $\times B$ )

= e(v₀i $\times B_0\mathrm{j}$ )=-eV₀B₀k

7. An electron mass m with an initial velocity v=v_oi is in an electric field E=-E_oi. its debroglie wavelength at time t is given by $\mathrm{T}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}.$

(a) $\frac{{\mathit{\lambda }}_{\mathit{o}}}{1+\frac{\mathit{e}{\mathit{E}}_{\mathit{o}}}{\mathit{m}}\frac{\mathit{t}}{{\mathit{v}}_{\mathit{o}}}}$

(b) ${\mathit{\lambda }}_{\mathit{o}}\left(1+\frac{\mathit{e}{\mathit{E}}_{\mathit{o}}\mathit{t}}{\mathit{m}{\mathit{v}}_{\mathit{o}}}\right)$

(c) ${\mathit{\lambda }}_{\mathit{o}}$

(d) ${\mathit{\lambda }}_{\mathit{o}}\mathit{t}$

Answer- a

Explanation- $\lambda =\frac{h}{mv}$

Force F= -eE=-e[E_oi]=eE_oi

So a=f/m=eE_oi/m

V=V_oi+ eE_oit/m

$\lambda =\frac{h}{mv}$ h/ m V_oi+ eE_oit/m

8. An electron mass m with initial velocity v=v₀i is an electric field E=E_o if $\lambda =\frac{h}{mv}=h/$ its debroglie wavelength at time t is given by

(a) ${\mathit{\lambda }}_{\mathit{o}}$

(b) ${\mathit{\lambda }}_{\mathit{o}}\sqrt[]{1+\frac{{\mathit{e}}^2{\mathit{E}}_{\mathit{o}}^2{\mathit{t}}^2}{{\mathit{m}}^2{\mathit{v}}_{\mathit{o}}^2}}$

(c) ${\mathit{\lambda }}_{\mathit{o}}/\mathit{}\sqrt[]{1+\frac{{\mathit{e}}^2{\mathit{E}}_{\mathit{o}}^2{\mathit{t}}^2}{{\mathit{m}}^2{\mathit{v}}_{\mathit{o}}^2}}$

(d) ${\mathit{\lambda }}_{\mathit{o}}/\mathit{}\sqrt[]{1+\frac{{\mathit{e}}^2{\mathit{E}}_{\mathit{o}}^2{\mathit{t}}^2}{{\mathit{m}}^2{\mathit{v}}_{\mathit{o}}^2}}$

Answer- c

Explanation- $\lambda =\frac{h}{mv}$

Force F= -eE=-e[E_oi]=eE_oi

So a=f/m=eE_oi/m

Velocity at xaxis=V_oi

Velocity at yaxis= eE_otj/m

Net velocity v= $\sqrt[]{v_x^2+v_y^2}=\sqrt[]{v_o^2+{\frac{\mathrm{e}\mathrm{E}\mathrm{o}\mathrm{t}}{\mathrm{m}}}^2}$

$\lambda =\frac{h}{mv}=h/m\sqrt[]{v_o^2+{\frac{\mathrm{e}\mathrm{E}\mathrm{o}\mathrm{t}}{\mathrm{m}}}^2}$

$\lambda ={\lambda }_o/\sqrt[]{1+\frac{e^2{E}_o^2{t}^2}{m^2{v}_o^2}}$

9. Relativistic corrections become necessary when the expression for the kinetic energy 1/2 mv², becomes comparable with mc². where m is the mass of the
At what de-Broglie wavelength, will relativistic corrections become important for an electron? 
(a) A=10nm (b) A =10^-1 nm (c) A=10^-4 nm (d) A=10^-6 nm

Answer-(c,d)

Explanation- $\lambda =\frac{h}{mv}$

V= h/m $\mathit{\lambda }$

V1=  $\lambda =10nm=10\times {10}^{-9}m={10}^{-8}$

$\frac{6.6\times {10}^{-34}}{9\times {10}^{-31}\times {10}^{-8}}={10}^{5}m/s$

V1=  $\lambda =1/10nm={10}^{-10}$

$\frac{6.6\times {10}^{-34}}{9\times {10}^{-31}\times {10}^{-10}}={10}^{7}m/s$

V1=  $\lambda =1/10000nm={10}^{-13}$

$\frac{6.6\times {10}^{-34}}{9\times {10}^{-31}\times {10}^{-13}}={10}^{10}m/s$

V1=  $\lambda =1/1000000nm={10}^{-15}$

10. Two particles A₁and A₂ of masses m₁ , m₂ ( m₁> m₂) have the same de-Broglie
Then,
(a) their momenta are the same

(b) their energies are the same
(c) energy of A₁ is less than the energy of A₂
(d) energy of A₁ is more than the energy of A₂

Answer-(a,c)

Explanation- as debroglie wavelength is $\lambda =\frac{h}{mv}=\frac{h}{p}$

P=h/ $\lambda$

P₁/p₂= ${\lambda }_2/{\lambda }_1$

If wavelength are equal then ratio is 1:1 so P₁=P₂

E= 1/2mv²= p²/2m

E inversely proportional to m

So $\frac{E_1}{E_2}$ = $\frac{m_2}{m_1}$ <1

E₁ 2

11. De-Broglie wavelength associated with uncharged particles: For Neutron efe-Broglie wavelength is given as v_e=c/100.Then

(a) $\frac{{\mathit{E}}_{\mathit{e}}}{{\mathit{E}}_{\mathit{p}}}$ = 10^-4 (b) $\frac{{\mathit{E}}_{\mathit{e}}}{{\mathit{E}}_{\mathit{p}}}$ = 10^-2 (c) $\frac{{\mathit{P}}_{\mathit{e}}}{{\mathit{m}}_{\mathit{e}}\mathit{c}}={10}^{-2}$ (d) $\frac{{\mathit{P}}_{\mathit{e}}}{{\mathit{m}}_{\mathit{e}}\mathit{c}}={10}^{-4}$

Answer- (b,c)

Explanation- ${\lambda }_e=\frac{h}{m_e{v}_e}$ = $\frac{h}{m_e\left(\frac{c}{100}\right)}=\frac{100h}{m_{e}c}$

E_e=1/2m_ev_e²

M_ev_e= $\sqrt[]{2E_e{m}_e}$

${\lambda }_e=\frac{h}{m_e{v}_e}=\frac{h}{\sqrt[]{2m_e{E}_e}}$

E_e= h²/2 $\lambda$ _e²m_e

For photon

E_p= hc/ ${\lambda }_p$ = hc/2 ${\lambda }_e$

$\frac{E_e}{E_p}=\frac{hc}{2{\lambda }_e}\times \frac{2{\lambda }_e^2{m}_e}{h^2}$ = 100

$\frac{E_e}{E_p}=\frac{1}{100}={10}^{-2}$

P_e=m_ev_e= m_{e $\times \frac{c}{100}$}

$\frac{P_e}{m_{e}c}=\frac{1}{100}={10}^{-2}$

(a) decreases with increasing n. with v fixed
(b) decreases with n fixed, v increasing
(c) remains constant with n and v changing such that nv = constant
(d) increases when the product nv increases

Answer-(a,b,c)

Explanation- energy spent to convert into water = mass $\times$ latent heat

= mL= 1000g $\times$ 80cal/g

= 80000cal

Energy of phtons used= nT $\times$ E=nT

So nTh $\times v$ =mL

T= mL/nhv

T $\propto$ 1/n where v is constant

T $\propto 1/v$ when n is fixed

T $\propto$ 1/nv

(c) When the de-Broglie wavetength is λ₁the particle is nearer the origin than when its value is λ₂
(d) When the de-Broglie wavelength is λ₂ the particle is nearer the origin than when its value is λ₁

Answer-(b,d)

Explanation- the debroglie wavelength of the particle can be varying cyclically between two values ${\lambda }_1$ and ${\lambda }_2$ , if particle is moving in an elliptical orbit with origin as its focus.

Let v₁ , v₂ , be the speed of particle at A and B respectively and origin is at focus O. If ${\lambda }_1$ and ${\lambda }_2$ are the de-Broglie wavelengths associated with particle while moving at A and B

respectively. Then,

${\lambda }_1$ =h/mv₁

${\lambda }_2$  =h/mv₂

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{v_2}{v_1}$

${\lambda }_1$ > ${\lambda }_2$

So v₂>v₁

By law of conservation of angular momentum, the particle moves faster when it is closer to

focus.

From figure, we note that origin O is closed to P than A