Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- from conservation of momentum

P_c=P_A+P_B

= $\frac{h}{{\lambda }_c}=\frac{h}{{\lambda }_A}+\frac{h}{{\lambda }_B}$

$\frac{h}{{\lambda }_c}=\frac{h{\lambda }_A+h{\lambda }_B}{{\lambda }_A{\lambda }_B}$

${\lambda }_c$ = $\frac{{\lambda }_A{\lambda }_B}{{\lambda }_A+{\lambda }_B}$

Case 1 when both momentum are positive

${\lambda }_c$ = $\frac{{\lambda }_A{\lambda }_B}{{\lambda }_A+{\lambda }_B}$

Case 2 when both momentum are negative

${\lambda }_c$ = $\frac{{\lambda }_A{\lambda }_B}{{\lambda }_A+{\lambda }_B}$

Case 3 when 1 is negative and 2 is positive

${\lambda }_c$ = $\frac{{\lambda }_A{\lambda }_B}{{\lambda }_B-{\lambda }_A}$

Case4 when 1 is positive and 2 is negative

${\lambda }_c$ = $\frac{{\lambda }_A{\lambda }_B}{{\lambda }_A-{\lambda }_B}$

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Explanation- Suppose nA is the number of photons falling per second of beam A and nB is the number of photons falling per second of beam B.

N_A=2n_B

energy of falling photon A=hv_A

B=hv_B

as we know intensities are same

n_Ahv_A=n_Bhv_B

v_a/v_b=n_B/n_A=1/2

v_B=2v_A

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Explanation-

$?p=\frac{h}{?x}=\frac{h}{2\pi ?x}$

= $\frac{6.64*{10}^{-34}}{2*22/7*{10}^{-9}}=1.05*{10}^{-25}$ kgm/s

E= p²/2m= $\frac{{(1.05*{10}^{-25})}^2}{2*9.1*{10}^{-31}}$ J= 3.8 $*{10}^{-2}$ eV

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Explanation- k'_max= 2k_max

K'_max= hc/ $\lambda -?$

2K_max= hc/ ${\lambda }^{\text{'}}-{?}_0$

2 (1230/600- $?$ )= (1230/400- $?$ )

So $?$ =1.02eV

This is a Short Answer Type Questions as classified in NCERT Exemplar

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- during photoelectric emission the momentum is transferred to metal. metal absorbs absorb the photon and its momentum is transferred, and excited electron emitted.

This is a Multiple Choice Questions as classified in NCERT Exemplar

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

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

= mL= 1000g $*$ 80cal/g

= 80000cal

Energy of phtons used= nT $*$ E=nT

So nTh $*v$ =mL

T= mL/nhv

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

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

T $\propto$ 1/nv

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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 ${\lambda }_e$

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

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

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

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

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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

E₁2