The graph of stopping potential V_s against f | 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:$e V_s = h 
u - \phi$$V_s = \frac{h}{e} 
u - \frac{\phi}

Vedclass · Accepted Answer

$\phi_x < \phi_y$

Vedclass · Answer

(B) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by:$e V_s = h 
u - \phi$$V_s = \frac{h}{e} 
u - \frac{\phi}{e}$Comparing this with the equation of a straight line $y = mx + c$,the intercept on the frequency axis (where $V_s = 0$) is the threshold frequency $
u_0$,given by $
u_0 = \frac{\phi}{h}$,which implies $\phi = h 
u_0$.From the graph,it is clear that the threshold frequency for metal $X$ $(
u_0)$ is less than the threshold frequency for metal $Y$ $(
u_0^{\prime})$,i.e.,$
u_0 Since the work function $\phi$ is directly proportional to the threshold frequency $
u_0$,we have $\phi_x < \phi_y$.

The graph of stopping potential $V_s$ against frequency $\nu$ of incident radiation is plotted for two different metals $X$ and $Y$ as shown in the graph. If $\phi_x$ and $\phi_y$ are the work functions of $X$ and $Y$,respectively,then:

Similar Questions

Radiations of wavelength $400 \,nm$ are incident on a photosensitive material with a work function of $2.2 \,eV$. The stopping potential is nearly: (in $\,V$)

The stopping potential as a function of frequency of incident radiation is plotted for two different photoelectric surfaces $A$ and $B$. The graph shows that the work function of $A$ is

$A$ material $A$ exhibits the photoelectric effect. Its work function is $2.5 \ eV$ and the threshold wavelength is $\lambda$. Another material $B$ has a work function of $5 \ eV$. Find the threshold wavelength required to produce the photoelectric effect in $B$.

When radiation of wavelength $400 \ nm$ is incident on a metal surface,the kinetic energy of the emitted photoelectrons is $1.68 \ eV$. The work function of the metal is .......... $eV$.

Light of two different frequencies whose photons have energies $1 \text{ eV}$ and $2.5 \text{ eV}$ respectively,successively illuminate a metallic surface whose work function is $0.5 \text{ eV}$. The ratio of the maximum speeds of the emitted electrons will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module