The variation of stopping potential for metal | vedclass

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

Question

(C) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by:$eV_s = h
u - \phi_o = h
u - h
u_o$$V_s = \frac{h}{e}(\

Vedclass · Accepted Answer

$A$

Vedclass · Answer

(C) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by:$eV_s = h
u - \phi_o = h
u - h
u_o$$V_s = \frac{h}{e}(
u - 
u_o)$Where $h$ is Planck's constant,$e$ is the charge of an electron,$
u$ is the frequency of incident radiation,and $
u_o$ is the threshold frequency.For a fixed incident frequency $
u$,the stopping potential $V_s$ is inversely proportional to the threshold frequency $
u_o$.From the graph,the threshold frequencies follow the order: $
u_o(A) Since metal $A$ has the lowest threshold frequency,it will have the highest stopping potential for any given incident frequency $
u$ (where $
u > 
u_o(D)$).Therefore,the correct option is $C$.

The variation of stopping potential for metals $A$,$B$,$C$ and $D$ with the frequency of incident radiation is shown in the figure. For which metal is the stopping potential higher for a given frequency of incident radiation $(v)$ if the threshold frequency is lower $(
u_o)$?

Similar Questions

In the graph given below,if the slope is $4.12 \times 10^{-15} \ V-sec$,then the value of $h$ should be:

Photoelectrons are emitted with maximum velocity $v$ when light of frequency $3f$ is incident on a photosensitive material of work function $2hf$. If the frequency of the incident light is $4.25f$,the maximum velocity of the emitted photoelectrons is ($h$ = Planck's constant).

When a photon of energy $8 \ eV$ is incident on a metal surface with a threshold frequency of $1.6 \times 10^{15} \ Hz$,the maximum kinetic energy of the emitted photoelectrons is .......... $eV$. (Given: $h = 6.6 \times 10^{-34} \ Js, 1 \ eV = 1.6 \times 10^{-19} \ J$)

The threshold wavelength of a metal surface is $5 \times 10^{-10} \, m$. When it is illuminated by light of wavelength $2 \times 10^{-10} \, m$,the stopping potential is $V_0$. What will be the stopping potential if the wavelength of the incident light is doubled?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The variation of stopping potential for metals $A$,$B$,$C$ and $D$ with the frequency of incident radiation is shown in the figure. For which metal is the stopping potential higher for a given frequency of incident radiation $(v)$ if the threshold frequency is lower $(u_o)$?