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)work function.

Answer 1

Correct option is A. (a) 4.14 X 10^{-15} eVs (b) 0.414 eV

We hav to take two cases :

Case I....$$v_0=1.656$$

$$v=5\times10^{14}Hz$$

Case II....$$v_0=0$$

$$v=1\times10^{14}Hz$$

We know ;

a) $$eV_0=h\nu-w_0$$

$$1.656e=h\times 5\times10^{-14}-w_0$$........(1)

$$0=5h\times10^{14}-5w_0$$ ........(2)

$$1.656e=4w_0$$

$$\Rightarrow w_0=\dfrac { 1.656 }{ 4 } ev=0.414ev$$

b) Putting value of $$w_0$$ in equation (2)

$$\Rightarrow5w_0=5h\times10^{14}$$

$$\Rightarrow5\times0.414=5\times h \times10^{14}$$

$$\Rightarrow h=4.414\times10^{-15}$$ ev-s