The energy of a photon with wavelength lambda | vedclass

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

Question

(D) According to Einstein's photoelectric equation: $\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi$. Given $\frac{hc}{\lambda} = 2 \ eV$,so $\frac{1}{2}

Vedclass · Accepted Answer

$1.8$

Vedclass · Answer

(D) According to Einstein's photoelectric equation: $\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi$. Given $\frac{hc}{\lambda} = 2 \ eV$,so $\frac{1}{2}mv^2 = 2 - \phi$  ---$(1)$ When $\lambda' = \lambda - 0.25\lambda = 0.75\lambda = \frac{3}{4}\lambda$ and $v' = 2v$,the new equation is: $\frac{1}{2}m(2v)^2 = \frac{hc}{0.75\lambda} - \phi$ $4(\frac{1}{2}mv^2) = \frac{4}{3}(\frac{hc}{\lambda}) - \phi$ Substituting $\frac{1}{2}mv^2 = 2 - \phi$ and $\frac{hc}{\lambda} = 2$: $4(2 - \phi) = \frac{4}{3}(2) - \phi$ $8 - 4\phi = \frac{8}{3} - \phi$ $8 - \frac{8}{3} = 3\phi$ $\frac{24 - 8}{3} = 3\phi$ $\frac{16}{3} = 3\phi$ $\phi = \frac{16}{9} \approx 1.777 \ eV \approx 1.8 \ eV$.

The energy of a photon with wavelength $\lambda$ is $2 \ eV$. When it strikes a metal surface,the maximum velocity of the emitted photoelectrons is $v$. If the value of $\lambda$ is decreased by $25\%$ and the maximum velocity is doubled,the work function of the metal becomes ...... $eV$.

Similar Questions

If the threshold wavelength for sodium is $6800 \mathring{A}$,then the work function will be ............. $eV$.

Light of wavelength $\lambda$ strikes a photo-sensitive surface and electrons are ejected with kinetic energy $E$. If the kinetic energy is to be increased to $2E$,the wavelength must be changed to $\lambda'$ where

When a metal is irradiated by light having wavelength $\lambda$ $(\lambda < \lambda_0)$,all the photoelectrons emitted are bent in a circle of radius $r$ by a magnetic field of flux density $B_0$. Find $\frac{1}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.

$A$ 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 $1 \ eV = 1.6 \times 10^{-19} \ J$)

In the graph of maximum kinetic energy of photoelectrons emitted from a metal surface versus the frequency of incident radiation based on Einstein's photoelectric equation,the slope is .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module