Question

The graph between $\frac{1}{\lambda }$ and stopping potential $\left(V\right)$ of three metals in an experiment of photo-electric effect is plotted as shown in the figure. The work function of metals $1,2,3$ are $W_1$, $W_2$ and $W_3$ respectively. Which of the following statement (s) is/are correct? [Here $\lambda$ is the wavelength of the incident ray]

• A. Ratio of work function $W_1:W_2:W_3=1:2:4$
• B. Ratio of work function $W_1:W_2:W_3=4:2:1$
• C. $\tan \theta$ is directly propotional to $\frac{hc}{e}$, where $h$ is planck's constant and $c$ is the speed of light.
• D. The violet colour light can eject photoelectrons from metals $2$ and $3$

Answer 1

Answer: (A), (C)

The correct options are
A Ratio of work function $W_1:W_2:W_3=1:2:4$
C $\tan \theta$ is directly propotional to $\frac{hc}{e}$, where $h$ is planck's constant and $c$ is the speed of light.

Work Function and threshold frequency are related as
$W=\frac{hc}{{\lambda }_0}\Rightarrow W_1:W_2:W_3=\frac{1}{{\lambda }_{10}}:\frac{1}{{\lambda }_{20}}:\frac{1}{{\lambda }_{30}}\Rightarrow W_1:W_2:W_3=0.001:0.002:0.004\Rightarrow W_1:W_2:W_3=1:2:4$

From Einstein's photo-electric equation, we have
$\frac{hc}{\lambda }=W+eV\Rightarrow V=\frac{hc}{e\lambda }-\frac{W}{e}$
comparing with,
$y=mx+c\Rightarrow m=\tan \theta =\frac{hc}{e}$
also,
${\lambda }_{20}=(0.002{)}^{-1}=500A^{\circ }{\lambda }_{30}=(0.004{)}^{-1}=250A^{\circ }$
To eject photoelectrons from 2 and 3 incident wavelength $\lambda <250A^{\circ }$
But wavelength of violet light ${\lambda }_{\text{violet}}\approx 4000A^{\circ }$
Thus the photoelectriceffect on metals 2 and 3 is not possible with violet light