Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

#School

When light of frequency $\nu _{1}$ is incident on a photosensitive surface, the stopping potential is V1. If the frequency of incident radiation becomes $\frac{\nu _{1}}{2},$ the stopping potential changes to V2. Find out the expression for the threshold frequency for the surface in terms of V1 and V2. If the frequency of incident radiation is doubled, will the maximum kinetic energy of the photoelectrons also be doubled ? Give reason.

Answers (1)

For the light of frequency $\nu _{1}$ and with stopping potential $V_{1}$ ,

$eV_{1}=h\nu _{1}-\phi$

Where h $\rightarrow$ Plank's constant

$\phi \rightarrow$ threshold frequency

For the light of frequency $\frac{\nu _{1}}{2}$ and stopping potential $V_{2}$ ,

$eV_{2}=h\frac{\nu _{1}}{2}-\phi$

$eV_{2}=\frac{h\nu _{1}-2\phi }{2}$

$2eV_{2}=h\nu _{1}-2\phi$

$h\nu _{1}=eV_{1}+\phi$ ______(1)

$h\nu _{1}=2eV_{2}+2\phi$ _____(2)

On equaling (1) and (2)

$eV_{1}+\phi=2eV_{2}+2\phi$

$eV_{1}-2eV_{2}=2\phi -\phi$

$\therefore \phi =e(V_{1}-2V_{2})$

$h\nu_0=e(V_1-2V_2)$

$\nu_0=\frac{1}{\nu_0}e(V_1-2V_2)$

If the frequency of incident radiation is doubled, the maximum kinetic energy of the photoelectrons will be more than double

$K_{max}=h\nu-h\nu_0$

When the frequency is doubled

$\\K_{max}_1=2h\nu-h\nu_0\\=2h\nu-h\nu_0-h\nu_0+h\nu_0\\=2KE_{max}+h\nu_0$

Posted by

Safeer PP

View full answer

Similar Questions

Latest Question