A photoemissive substance is illuminated with | 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 kinetic energy $K$ of the emitted photoelectron is given by $K = \frac{hc}{\lambda_i} - \frac{

Vedclass · Accepted Answer

$[h/[2mc(\frac{1}{\lambda_i} - \frac{1}{\lambda_0})]]^{1/2}$

Vedclass · Answer

(D) According to Einstein's photoelectric equation, the kinetic energy $K$ of the emitted photoelectron is given by $K = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0}$.Substituting $K = \frac{p^2}{2m}$, where $p$ is the momentum of the electron, we get $\frac{p^2}{2m} = hc(\frac{1}{\lambda_i} - \frac{1}{\lambda_0})$.From the de-Broglie relation, the wavelength $\lambda_e$ is given by $\lambda_e = \frac{h}{p}$, which implies $p = \frac{h}{\lambda_e}$.Substituting $p$ in the kinetic energy equation: $\frac{h^2}{2m\lambda_e^2} = hc(\frac{1}{\lambda_i} - \frac{1}{\lambda_0})$.Rearranging for $\lambda_e^2$: $\lambda_e^2 = \frac{h^2}{2mhc(\frac{1}{\lambda_i} - \frac{1}{\lambda_0})} = \frac{h}{2mc(\frac{1}{\lambda_i} - \frac{1}{\lambda_0})}$.Thus, $\lambda_e = [\frac{h}{2mc(\frac{1}{\lambda_i} - \frac{1}{\lambda_0})}]^{1/2}$.

Similar Questions

According to Einstein's photoelectric equation,the graph of the maximum kinetic energy of emitted photoelectrons versus the frequency of incident radiation is a straight line. Its slope . . . . . .

When a photosensitive surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $V/3$. The threshold wavelength for the surface is:

In the photoelectric effect,the kinetic energy $(K.E.)$ of electrons emitted from the metal surface depends upon:

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