Question

The graph between $1/\lambda$ and stopping potential (V) of three metals having work functions ${\varphi }_1,{\varphi }_2$ and ${\varphi }_3$ in an experiment of photoelectric effect is plotted as shown in the figure. Which of the following statement(s) is/are correct ?
[Here $\lambda$ is the wavelength of the incident ray].

• A. Ratio of work functions ${\varphi }_1:{\varphi }_2:{\varphi }_3=1:2:4.$
• B. Ratio of work functions ${\varphi }_1:{\varphi }_2:{\varphi }_3=4:2:1.$
• C. tan $\theta$ is directly proportional to $hc/e$, where $h$ is Plancks 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 functions ${\varphi }_1:{\varphi }_2:{\varphi }_3=1:2:4.$
C tan $\theta$ is directly proportional to $hc/e$, where $h$ is Plancks constant and $c$ is the speed of light.

According to Einstein's equation, $\frac{hc}{\lambda }-\varphi =$ eV
or, $V=\frac{hc}{e\lambda }-\frac{\varphi }{e}$
Now the ratio, plate 1 : Plate 2 : Plate 3
$\frac{{\varphi }_1}{e}=0.001$, $\frac{{\varphi }_2}{e}=0.002$, $\frac{{\varphi }_3}{e}=0.004$

${\varphi }_1:{\varphi }_2:{\varphi }_3=1:2:3$
For plate 2, threshold wavelength, $\lambda =\frac{hc}{{\varphi }_2}=\frac{hc}{0.002hc}=\frac{1000}{2}=500nm$

For plate 3, threshold wavelength, $\lambda =\frac{hc}{{\varphi }_3}=\frac{hc}{0.004hc}=\frac{1000}{4}=$250nm
Since violet colour light $\lambda$ is 400 nm, so ${\lambda }_{violet}$ < ${\lambda }_{threshold}$ for plate 2.
So, violet colour light will eject photoelectrons from plate 2 and not from plate 3.