The stopping potential required to reduce the | vedclass

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

Question

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ of emitted photoelectrons is given by: $K_{max} = h
u - \Ph

Vedclass · Accepted Answer

increases linearly with the frequency of the incident light.

Vedclass · Answer

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ of emitted photoelectrons is given by: $K_{max} = h
u - \Phi$,where $h$ is Planck's constant,$
u$ is the frequency of incident light,and $\Phi$ is the work function of the metal.Since the stopping potential $(V_s)$ is related to the maximum kinetic energy by $K_{max} = eV_s$,we have: $eV_s = h
u - \Phi$.Rearranging for $V_s$,we get: $V_s = (h/e)
u - (\Phi/e)$.This equation represents a linear relationship between the stopping potential $(V_s)$ and the frequency $(
u)$ of the incident light,where the slope is $(h/e)$ and the intercept is $-(\Phi/e)$.Therefore,the stopping potential increases linearly with the frequency of the incident light.

The stopping potential required to reduce the photoelectric current to zero is . . . . . .

Similar Questions

Find the number of electrons emitted per second by a $24 \, W$ source of monochromatic light of wavelength $6600 \, \mathring{A}$, assuming $3 \%$ efficiency for the photoelectric effect (take $h = 6.6 \times 10^{-34} \, J \cdot s$ and $c = 3 \times 10^8 \, m/s$).

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

$A$ photon of energy $8 \ eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15} \ Hz$. The maximum kinetic energy of the photoelectrons emitted (in $eV$) is: (Take $h = 6 \times 10^{-34} \ J \cdot s$ and $1 \ eV = 1.6 \times 10^{-19} \ J$)

$A$ metal surface of work function $1.07 eV$ is irradiated with light of wavelength $332 nm$. The retarding potential required to stop the escape of photo-electrons is .............. $V$.

When radiation of the wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then,the threshold wavelength for the surface is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module