Question

The figure is the plot of stopping potential versus the frequency of the light used in an experiment on photoelectric effect. Find (a) the ratio h/e and (b) the work function.
Figure

Answer 1

(a) The ratio $\frac{h}{e}$ is $4.414 \times 10^{-15} \mathrm{Vs}$

(b) The work function is $0.414 \mathrm{eV}$

Explanation:

We've got 2 cases to handle.

Case (I):- When potential  stopping  , $V_0$ =1.656 volts

Frequency, $v=5 \times 10^{14} \mathrm{Hz}$

Case (II):- When potential  stopping , $V_0$ =0

Frequency, $v=1 \times 10^{14} \mathrm{Hz}$

(b)To find: The work function.

From the Equation of Einstein,  $e V_{0}=h v-W_{0}$

When the case(1) and case(2) values are replaced we get:

$\begin{aligned}&1.656 e=h \times 5 \times 10^{14}-W_{0} \ldots(1)\\&0=5 \times h \times 1 \times 10^{14}-5 \times W_{0} \ldots(2)\end{aligned}$

We get: Subtract from (1) equation(2)

$W_{0}=\frac{1.656}{4} e V$    

     = 0.414 eV

(a)To find the ratio of  $\frac{h}{e}$ :

Putting the $W_0$ value into equation (2), we get:

$5 W_{0}=5 h \times 10^{14}$

$5 \times 0.414=5 \times h \times 10^{14}$

$h=4.414 \times 10^{-15} \mathrm{eVs}$

Or  

$\frac{h}{e}=4.414 \times 10^{-15} \mathrm{Vs}$