fig. shows variation of stopping potential `(V_0)` with frequency (v) for two photosenstive matrials `M_1 and M_2`.

(i) why is the slope same for both lines?
(ii) For which material will the emitted electrons have greater kinetic energy for the incident radiations of the same frequency? Justify your answer.

Explanation

(i) From photoelectric equation, we have
`eV_0-hv-phi_0 or V_0=(hv)/e-(phi_0)/e`
It is an equation of straight line with slope h/e (=constant). It means the slope metals `V_0-v` graph
(=h/e) is same for both metals `M_1 and M_2`.
(ii) Also, `K_(max)=hv-hv_0`. For the given frequency of incident light, the smaller is the value of `v_0`, the larger is the value of `K_(max)` and vice versa. Since material `M_1` has lower value of threshold frequency `v_0`, so metal `M_1` will emit photoelectrons of greater K.E.