When radiation of wavelength lambda is incide | vedclass

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

Question

(D) According to Einstein's photoelectric equation: $\frac{1}{2}m\upsilon^2 = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the wo

Vedclass · Accepted Answer

$ > \,\,\upsilon \,{\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$

Vedclass · Answer

(D) According to Einstein's photoelectric equation: $\frac{1}{2}m\upsilon^2 = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function.Thus,$\upsilon = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} ight)} = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} ight)} \dots (1)$When the wavelength is changed to $\lambda' = \frac{3\lambda}{4}$,the new velocity $\upsilon'$ is:$\upsilon' = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - 3\lambda/4}{(3\lambda/4) \lambda_0} ight)} \dots (2)$Dividing equation $(2)$ by $(1)$:$\frac{\upsilon'}{\upsilon} = \sqrt{\frac{\lambda_0 - 3\lambda/4}{3\lambda/4 \lambda_0} 	imes \frac{\lambda \lambda_0}{\lambda_0 - \lambda}} = \sqrt{\frac{4}{3} \left( \frac{\lambda_0 - 3\lambda/4}{\lambda_0 - \lambda} ight)}$Since $\lambda_0 - 3\lambda/4 > \lambda_0 - \lambda$,the term $\frac{\lambda_0 - 3\lambda/4}{\lambda_0 - \lambda} > 1$.Therefore,$\upsilon' > \upsilon \left( \frac{4}{3} ight)^{1/2}$.

When radiation of wavelength $\lambda$ is incident on a photoelectric cell,the maximum velocity of the emitted photoelectrons is $\upsilon$. If the wavelength of the incident radiation is changed to $3\lambda / 4$,the maximum velocity of the emitted photoelectrons will be ..........

Similar Questions

The work function of a metal is $6.825 \ eV$. Its threshold wavelength is approximately: (Given $c = 3 \times 10^8 \ m/s$)

The retarding potential necessary to stop the emission of photoelectrons,when a target material of work function $1.24 eV$ is irradiated with light of wavelength $4.36 \times 10^{-7} m$ is (in $eV$)

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light $(v_1 > v_2)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1: n$,then the threshold frequency of the metallic surface is

The graph of maximum kinetic energy $(K.E._{max})$ against the frequency $(\nu)$ of incident light is as shown in the figure. The slope of the graph and the intercept on the $X$-axis respectively are:

If the wavelengths of incident radiation are $2500 \ \mathring A$ and $5000 \ \mathring A$ respectively,and the work function of the metal surface is $2 \ eV$,find the approximate ratio of the stopping potentials for the emitted photoelectrons. (in $:1$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module