Question

The graph between the stopping potential ($V_0$) and $\left(\frac{1}{\lambda }\right)$ is shown in the figure. ${\varphi }_1$,${\varphi }_2$ and ${\varphi }_3$ are work functions, which of the following is/are correct.

• A.
• B.
• C.
• D. I don't want to do this, it looks too tough.

Answer 1

Answer: (A), (C)

The correct options are
A

C

The graph is between stopping potential and $\frac{1}{\lambda }$ . If we can find a relation between them then we can solve this. So let’s go ahead.

$\lambda$ is for the incident light whereas $V_0$ is the stopping potential. How do we find $V_0$ . Let’s recall If photon of energy E or $\frac{hc}{\lambda }$ falls on electron it uses φ Joules of energy to overcome the work function and rest is converted to its kinetic energy (K.E).

For stopping potential ($V_0$)

$e{V}_0=K{E}_{max}$

$\Rightarrow e{V}_0=\frac{hc}{\lambda }-\varphi$

$\Rightarrow V_0=\frac{hc}{e\lambda }-\left(\frac{\varphi }{e}\right)$

Slope is a constant

From the graph we can see that the x intercept for metal 1 < metal 2 < metal 3

Well if you forgot how to find the x intercept then let me refersh it for you. It’s basically the x coordinate when y is ‘o’ If you substitute that in equation we get $\left(\frac{hc}{\lambda }\right)=\varphi$

So the x intercept = $\frac{\varphi }{hc}$

Given Metal 1 < Metal 2 < Metal 3

$\Rightarrow \left(\frac{{\varphi }_1}{hc}\right)<\left(\frac{{\varphi }_2}{hc}\right)<\left(\frac{{\varphi }_3}{hc}\right)$

$\Rightarrow {\varphi }_1<{\varphi }_2<{\varphi }_3$

Also ratio of x intercept of metals

Metal 1 : Metal 2 : Metal 3

0.001: 0.002 : 0.004

1: 2 : 4

So this will be equal to the ratio of work functions too

${\varphi }_1$: ${\varphi }_2$ :${\varphi }_3$

1 : 2 : 4

So option (a) is correct, (b) is wrong.

In a straight line graph the slope of it is given by the tan of the angle that it makes wilh the x – axis. Also in y = mx + c form ‘m’ or coefficient of x is the slope. The equation of our line is

$V_0=\frac{hc}{e\lambda }-\left(\frac{\varphi }{e}\right)$

Clearly the slope is $tan\theta$ or $\left(\frac{hc}{e}\right)$