When a metallic surface is illuminated with l | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $

Vedclass · Accepted Answer

$4 \lambda$

Vedclass · Answer

(B) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.For the first case: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$  --- $(1)$For the second case: $e(\frac{V}{3}) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$  --- $(2)$Multiply equation $(2)$ by $3$: $eV = \frac{3hc}{2\lambda} - \frac{3hc}{\lambda_0}$  --- $(3)$Equating $(1)$ and $(3)$: $\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{3hc}{2\lambda} - \frac{3hc}{\lambda_0}$Rearranging the terms: $\frac{3hc}{\lambda_0} - \frac{hc}{\lambda_0} = \frac{3hc}{2\lambda} - \frac{hc}{\lambda}$$\frac{2hc}{\lambda_0} = \frac{hc}{2\lambda}$$\frac{2}{\lambda_0} = \frac{1}{2\lambda}$$\lambda_0 = 4\lambda$.

When a metallic surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential is $\frac{V}{3}$. The threshold wavelength for the metallic surface is:

Similar Questions

In the photoelectric effect,if the intensity of light is doubled,then the maximum kinetic energy of the photoelectrons will become:

When a photosensitive material is illuminated by photons of energy $3.1 \ eV$,the stopping potential of the photoelectrons is $1.7 \ V$. When the same photosensitive material is illuminated by photons of energy $2.5 \ eV$,the stopping potential of the photoelectrons is: (in $V$)

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. If the maximum kinetic energies of the photoelectrons emitted in the two cases are in the ratio of $1:k$, then the threshold frequency of the metallic surface is:

Which one of the following graphs correctly represents the variation of maximum kinetic energy $(E_{k})$ of the emitted electrons with frequency $(\nu)$ of incident light in the photoelectric effect?

Light of wavelength $\lambda$ strikes a photoelectric surface and electrons are ejected with kinetic energy $K$. If $K$ is to be increased to twice its original value,the wavelength must be changed to $\lambda'$ such that :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module