CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 8 - MCQExams.com

Let initial momentum is $${{p}_{i}}$$and final reflected momentum is $${{p}_{f}}$$such that

Energy, $$E=\dfrac{hc}{\lambda }.....(1)$$

$$ {{p}_{i}}=\dfrac{h}{\lambda }=\dfrac{E}{c}\,\,(from\,\,1) $$

$$ {{p}_{f}}=\dfrac{-h}{\lambda }=\dfrac{-E}{c} $$

So, net change in momentum is

$$ \Delta p={{p}_{f}}-{{p}_{i}} $$

$$ \Delta p=\dfrac{-E}{c}-\dfrac{E}{c} $$

$$ \Delta p=-\dfrac{2E}{c} $$

The momentum transferred to the surface is$$\dfrac{2E}{c}$$

We Know,
According to De-Broglie equation: 

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

$$p= \dfrac{h\times c}{\lambda \times c}= \dfrac{\textrm{energy}}{c}$$

$$p= \dfrac{9\times 1.6\times 10^{-19}}{3\times 10^8}$$

$$p=\dfrac{14.4 \times 10^{-27}}{3}$$

$$p=4.8\times 10^{-27}$$

Hence option $$\textbf B$$ is correct.

Given that,

The maximum wave length $$\lambda_{0} =200\,nm$$

Radiation of wave length $$\lambda =100\,nm$$

We know that, the maximum K.E is

  $$ K.E=12400\left[ \dfrac{1}{\lambda }-\dfrac{1}{{{\lambda }_{0}}} \right] $$

 $$ K.E=12400\left[ \dfrac{1}{1000}-\dfrac{1}{2000} \right] $$

 $$ K.E=12400\times \dfrac{1}{2000} $$

 $$ K.E=6.2\,eV $$

Hence, the kinetic energy is $$6.2\ eV$$

Given that,

$$T=20+273=293\,K$$

Molecular mass of $${{H}_{2}}=2\times 1.00794=2.01588\times 1.66\times {{10}^{-27}}\,kg$$

We know that,

  $$ \dfrac{1}{2}mv_{rms}^{2}=\dfrac{3}{2}kT $$

 $$ {{v}_{rms}}=\sqrt{\dfrac{3kt}{m}} $$

Now, the de Broglie wave length

  $$ \lambda =\dfrac{h}{m{{v}_{rms}}} $$

 $$ \lambda =\dfrac{h}{\sqrt{3mkT}} $$

 $$ \lambda =\dfrac{6.63\times {{10}^{-34}}}{\sqrt{3\times 2.01588\times 1.66\times {{10}^{-27}}\times 1.38\times {{10}^{-23}}\times 293}} $$

 $$ \lambda =\dfrac{6.63\times {{10}^{-34}}}{\sqrt{4059.20\times {{10}^{-50}}}} $$

 $$ \lambda =\dfrac{6.63\times {{10}^{-34}}}{63.712\times {{10}^{-25}}} $$

 $$ \lambda =0.104\times {{10}^{-9}} $$

 $$ \lambda =1.04\overset{\circ }{\mathop{A}}\, $$

Hence, the de Broglie wave length is $$1.04\overset{\circ }{\mathop{A}}\,$$ 

Kinetic energy of photoelectron is $${{E}_{1}}$$ and $${{E}_{2}}$$ for incident light of frequency $$\upsilon $$ and $$2\upsilon $$ respectively

Now, $$h\upsilon ={{E}_{1}}+{{\phi }_{0}}.....(I)$$

$$2h\upsilon ={{E}_{2}}+{{\phi }_{0}}.....(II)$$

Now, put the value of $$h\upsilon $$ in equation (II)

  $$ 2\left( {{E}_{1}}+{{\phi }_{o}} \right)={{E}_{2}}+{{\phi }_{0}} $$

 $$ {{E}_{2}}=2{{E}_{1}}+{{\phi }_{0}} $$

So, the kinetic energy of the emitted photoelectrons becomes more then two times its initial value

$$K = \displaystyle {{hc} \over \alpha } - \phi $$

$$ = \displaystyle {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {200 \times {{10}^{ - 9}} \times 1.6 \times {{10}^{ - 19}}}} - 4.5$$

$$ = 1.7eV$$

Min. energy with which electrons emitted=1.7eV

Max. $$K\varepsilon  \Rightarrow {K_{\min }} + e{V_0}$$

$$ = 1.7 + 2$$

$$ = 3.7eV$$

De Broglie wavelength in terms of accelerated potential difference is

$$\lambda =\dfrac{12.27}{\sqrt{V}}$$

Initial and final wavelength are $${{\lambda }_{1}}$$and $${{\lambda }_{2}}$$having potential as $${{V}_{1}}$$and $${{V}_{2}}$$

$$ \dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\sqrt{\dfrac{{{V}_{2}}}{{{V}_{1}}}} $$

$$ \dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\sqrt{\dfrac{100}{25}} $$

$$ \dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\sqrt{2} $$

$$ \dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=2 $$

$$ {{\lambda }_{2}}=\dfrac{{{\lambda }_{1}}}{2} $$

So, new wavelength is decrease by 2.