Dual Nature of Radiation and Matter

Case I :- $2?-?=\frac{1}{2}m{v}_1^2........\left(i\right)$  

Case II :-     $10?-?=\frac{1}{2}m{v}_2^2........\left(ii\right)$

$\Rightarrow \frac{1}{9}=\frac{v_1^2}{v_2^2}$                                        

$\Rightarrow v_1:v_2=1:3$                            

->x = 1

${\lambda }_e=\frac{h}{mv}=\frac{h}{p_e}$    -(1)

${\lambda }_{Ph}=\frac{h}{P_{Ph}}$              -(2)

A/C to question

${\lambda }_e={\lambda }_{Ph}$  

$P_e=P_{Ph}\Rightarrow \frac{P_P}{P_{Ph}}=1$              -(3)

$k.E_e=E_e=\frac{1}{2}m{v}^2$  

$=\frac{1}{2}Pev$              -(4)

$k.E_{Ph}=E_{Ph}=m{c}^2$  

   = mc c

q = P_Ph c               -(5)

$\left(4\right)÷\left(5\right)$  

$\frac{E_e}{E_{Ph}}=\frac{v}{2c}$  

The light wave contains two lights of different frequencies, so

  $E_1=h{\upsilon }_1=\frac{4.14*10^{-15}*6*10^{15}}{2\pi }=3.96eV,and$

  $E_2=h{\upsilon }_2=\frac{4.14*10^{-15}*9*10^{15}}{2\pi }=5.92eV$

Maximum kinetic energy of the photoelectron = 5.9-2.5 = 3.42 eV

The light wave contains two lights of different frequencies, so

$E_1=h{\upsilon }_1=\frac{4.14*10^{-15}*6*10^{15}}{2\pi }=3.96eV,and$  

$E_2=h{\upsilon }_2=\frac{4.14*10^{-15}*9*10^{15}}{2\pi }=5.92eV$  

Maximum kinetic energy of the photoelectron = 5.9-2.5 = 3.42 eV

${\lambda }_1=\frac{h}{\sqrt[]{2m{E}_1}}={\lambda }_2=\frac{h}{\sqrt[]{2m{E}_2}}$

$=E_2=\frac{4}{9}{E}_1=4eV$

$E_2=E_1-e{V}_0=V_0=5V$

This experiment is the famous Davission-Germer experiment which establishes wave nature of electrons.

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

${\lambda }_p={\lambda }_{\alpha }\left[Given\right]$

$\frac{h}{m_p{V}_p}=\frac{h}{m_{\alpha }{v}_{\alpha }}\Rightarrow \frac{v_p}{v_{\alpha }}=\frac{m_{\alpha }}{m_p}$

$\Rightarrow \frac{v_p}{v_{\alpha }}=\frac{4m}{m}=\frac{4}{1}$ or 4 : 1

Intersity a number of photons kinetic Energy a f

Photon energy $\frac{}{}\frac{}{}{}^{}{}^{}$ $=\frac{hc}{\lambda }=\frac{1240\mathrm{e}\mathrm{V}.\mathrm{n}\mathrm{m}}{500\mathrm{n}\mathrm{m}}=2.48\mathrm{e}\mathrm{V}=2.48*1.6*{10}^{-19}\mathrm{J}=4*{10}^{-19}\mathrm{J}$

No. of photons emitted per sec $=\frac{\text{power}}{\text{photon Energy}}=\frac{100}{4*{10}^{-19}}{s}^{-1}=2.5*{10}^{20}{s}^{-1}$

Stopping potential increases and number of photons decreases.

$I=\frac{hc{n}_p}{\lambda t}$