Question

The graph between $1 / \lambda$ and stopping potential $(V)$ of three metals having work functions $\phi_{1}, \phi_{2}$ and $\phi_{3}$ in an experiment of photoelectric effect is plotted as shown in the figure. Which of the following option(s) is/are correct?
(Here, $\lambda$ is the wavelength of the incident ray)

• A. Ratio of work function $\phi_{1}: \phi_{2}: \phi_{3}=1: 2: 4$
• B. Ratio of work function $\phi_{1}: \phi_{2}: \phi_{3}=4: 2: 1$
• C. $\tan \theta$ is directly proportional to $h c / e$, 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)

$\frac{h c}{\lambda}-\phi=e V$
$V=\frac{h c}{e \lambda}-\frac{\phi}{e}$
For plate I
$\frac{\phi_{1}}{h c}=0.001$
Plate 2
$\frac{\phi_{2}}{h c}=0.002$
Plate 3
$\frac{\phi_{3}}{h c}=0.004$
$\phi_{1}: \phi_{2}: \phi_{3}=1: 2: 4$
For plate $2$ , threshold wavelength
$\lambda=\frac{h c}{\phi_{2}}=\frac{h c}{0.002 h c}=\frac{1000}{2}=500\, nm$
For plate $3$ , threshold wavelength
$\gamma=\frac{h c}{\phi_{3}}=\frac{h c}{0.004 h c}=\frac{1000}{4}=250\, nm$
Since, violet colour light will eject photoelectrons from plate 2$ not from $3$ .