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 photo-electric 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 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 functions ${\varphi }_1$ : ${\varphi }_2$ : ${\varphi }_3=1$ : 2 : 4
C $\tan \Theta$ is directly proportional to $hc/e$, where $h$ is Planck's constant and $c$ is the speed of light.

From photo electric effect, $eV=\frac{hc}{\lambda }-\varphi$ where $V=$ stopping potential and $\varphi =$ work function.

Slope , $m=\tan \theta =hc/e$ (compare above equation with $y=mx-c$)

If $V=0\Rightarrow \frac{\varphi }{hc}=\frac{1}{\lambda }$

Now from given graph,

for metal 1 : $\frac{{\varphi }_1}{hc}=0.001$

for metal 2: $\frac{{\varphi }_2}{hc}=0.002$

for metal 3: $\frac{{\varphi }_3}{hc}=0.004$

Thus, ${\varphi }_1:{\varphi }_2:{\varphi }_3=1:2:4$

Also, $\varphi =\frac{hc}{{\lambda }_0}$ where ${\lambda }_0$ be the threshold wavelength.

For metal 2 : ${\lambda }_{02}=1/0.002=500nm$

For metal 3 : ${\lambda }_{02}=1/0.004=250nm$

As the wavelength of violet colour is around 400 nm so only metal 2 can eject violet colour light but not metal 3.

Thus, options A and C is right.