Question

Figure (42-E2) is the plot of the 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.

Answer 1

We have to take two cases

Case-I $V_c=1.656$

$v=5\times {10}^{14}\text{Hz}$

Case-II $v_0=0$

$v=1\times {10}^{14}\text{Hz}$

(b) We know $e{v}_0=hv-W_0$

$1.656e=h\times 5\times {10}^{14}-W_0$

$0=5h\times {10}^{14}-5W_0...\left(1\right)$

$1.6556e=4W_0...\left(2\right)$

$\Rightarrow W_0=\frac{1.656}{4}eV$

$=0.414eV$

(a) Putting the value of $W_0$ in equation (2)

$\Rightarrow 5W_0=5h\times {10}^{14}$

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

$\Rightarrow h=4.414\times {10}^{-15}eV-s$