A photo-emissive substance is illuminated wit | vedclass

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

Question

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E.)$ of the emitted photoelectrons is given by: $K.E. = E - W$,where

Vedclass · Accepted Answer

$\lambda_c = \sqrt{\frac{h}{2mc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)}}$

Vedclass · Answer

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E.)$ of the emitted photoelectrons is given by: $K.E. = E - W$,where $E = \frac{hc}{\lambda_i}$ is the energy of the incident photon and $W = \frac{hc}{\lambda_0}$ is the work function of the substance.Substituting these,we get: $K.E. = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0} = hc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)$.The de-Broglie wavelength $\lambda_c$ of an electron with kinetic energy $K.E.$ is given by $\lambda_c = \frac{h}{\sqrt{2m(K.E.)}}$.Squaring both sides,we get $\lambda_c^2 = \frac{h^2}{2m(K.E.)}$,which implies $K.E. = \frac{h^2}{2m\lambda_c^2}$.Equating the two expressions for $K.E.$: $\frac{h^2}{2m\lambda_c^2} = hc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)$.Solving for $\lambda_c$: $\lambda_c^2 = \frac{h^2}{2mhc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)} = \frac{h}{2mc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)}$.Therefore,$\lambda_c = \sqrt{\frac{h}{2mc \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)}}$.

Similar Questions

When photons of energy $1 \ eV$ and $2.5 \ eV$ are incident on a metal surface with a work function of $0.5 \ eV$,what is the ratio of the maximum kinetic energies of the emitted photoelectrons?

$A$ 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 light of wavelength $\lambda$ is incident on a photosensitive surface,the stopping potential is $V$. When light of wavelength $3 \lambda$ is incident on the same surface,the stopping potential is $\frac{V}{6}$. Then the threshold wavelength for the surface is:

Two identical photocathodes receive light of frequencies $n_1$ and $n_2$. If the velocities of the emitted photoelectrons of mass $m$ are $V_1$ and $V_2$ respectively, then ($h$ = Planck's constant)

When light of frequency $8 \times 10^{14} \ Hz$ is incident on a surface,photoelectrons are emitted with a maximum speed of $7 \times 10^5 \ m/s$. The threshold frequency for this surface is ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module