When a photosensitive metal surface is illumi | vedclass

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

Question

(C) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by: $eV_s = \frac{hc}{\lambda} - \Phi$,where $\Phi = \frac{hc

Vedclass · Accepted Answer

$5 \lambda_1$

Vedclass · Answer

(C) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by: $eV_s = \frac{hc}{\lambda} - \Phi$,where $\Phi = \frac{hc}{\lambda_0}$ is the work function and $\lambda_0$ is the threshold wavelength.For the first case: $eV_1 = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_0} = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_0} \right)$  --- $(1)$For the second case: $e \left( \frac{V_1}{6} \right) = \frac{hc}{3\lambda_1} - \frac{hc}{\lambda_0} = hc \left( \frac{1}{3\lambda_1} - \frac{1}{\lambda_0} \right)$ --- $(2)$Dividing equation $(1)$ by equation $(2)$:$6 = \frac{\frac{1}{\lambda_1} - \frac{1}{\lambda_0}}{\frac{1}{3\lambda_1} - \frac{1}{\lambda_0}}$$6 \left( \frac{1}{3\lambda_1} - \frac{1}{\lambda_0} \right) = \frac{1}{\lambda_1} - \frac{1}{\lambda_0}$$\frac{2}{\lambda_1} - \frac{6}{\lambda_0} = \frac{1}{\lambda_1} - \frac{1}{\lambda_0}$$\frac{2}{\lambda_1} - \frac{1}{\lambda_1} = \frac{6}{\lambda_0} - \frac{1}{\lambda_0}$$\frac{1}{\lambda_1} = \frac{5}{\lambda_0}$$\lambda_0 = 5\lambda_1$. Thus,the correct option is $C$.

Similar Questions

In the photoelectric effect,a graph is plotted between the maximum kinetic energy of electrons and the frequency of incident photons. The slope of this curve will be .......

The figure shows different graphs between stopping potential $(V_0)$ and frequency $(\nu)$ for photosensitive surfaces of cesium,potassium,sodium,and lithium. The plots are parallel. The correct ranking of the targets according to their work function,with the greatest first,is:

In the graph given below,if the slope is $4.12 \times 10^{-15} \, V-s$,then the value of $'h'$ should be:

The photoelectric threshold wavelength for silver is $\lambda_{0}$. The energy of the electron ejected from the surface of silver by an incident wavelength $\lambda$ (where $\lambda < \lambda_{0}$) will be:

If the work function for a certain metal is $3.2 \times 10^{-19} \ J$ and it is illuminated with light of frequency $8 \times 10^{14} \ Hz$,the maximum kinetic energy of the photo-electrons would be (given $h = 6.63 \times 10^{-34} \ J \cdot s$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module