The stopping potential V_0 (in volt) as a fun | vedclass

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

Question

(B) According to Einstein's photoelectric equation:$h
u = \phi + eV_0$Rearranging for the stopping potential $V_0$:$V_0 = \frac{h
u}{e} - \frac{\phi

Vedclass · Accepted Answer

$1.66$

Vedclass · Answer

(B) According to Einstein's photoelectric equation:$h
u = \phi + eV_0$Rearranging for the stopping potential $V_0$:$V_0 = \frac{h
u}{e} - \frac{\phi}{e}$From the given graph,the threshold frequency $(
u_0)$ is the frequency at which the stopping potential $V_0$ becomes zero. Looking at the graph,$V_0 = 0$ when $
u = 4 	imes 10^{14} \, Hz$.At this point:$0 = \frac{h
u_0}{e} - \frac{\phi}{e}$$\Rightarrow \phi = h
u_0$Substituting the values:$\phi = (6.63 	imes 10^{-34} \, Js) 	imes (4 	imes 10^{14} \, Hz)$$\phi = 26.52 	imes 10^{-20} \, J$To convert the work function into electron-volts $(eV)$:$\phi (eV) = \frac{26.52 	imes 10^{-20} \, J}{1.6 	imes 10^{-19} \, C}$$\phi = 1.6575 \, eV \approx 1.66 \, eV$

Similar Questions

Albert Einstein was awarded the Nobel Prize in Physics for his work on

When the energy of the incident radiation is increased by $20\%$,the kinetic energy of the photoelectrons emitted from a metal surface increases from $0.5\, eV$ to $0.8\, eV$. The work function of the metal is ............. $eV$.

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 the wavelength of an incident photon is decreased, then:

The wave equation of an electric field at a point is $E = 100 \frac{V}{m} [\sin(7 \omega t) + \cos(10 \omega t) + \cos(15 \omega t)]$ at instant $t$. If the work function of the photocell is $\phi$,then the stopping potential is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module