Question

For a constant frequency of the incident light, the graph between frequency (ν) of the incident light, and the magnitude (V_s), of the stopping potential, for metal A, is as shown. Light of frequency (nν) (n > 1), is now made to fall, one by one, on metal A and another metal B. The magnitude of stopping potential, for this frequency for metal B, equal V₀ . The (corresponding) magnitude of the stopping potential, for metal A, and the threshold frequency, for metal B, are equal, respectively, to

(1) \([(n-1)\frac{hv_0}{e}]\) and \([(n-1)v_0-\frac{eV_0}{h}]\)

(2) \([(n-1)\frac{hv_0}{e}]\) and \([nv_0-\frac{eV_0}{h}]\)

(3) \([n\frac{hv_0}{e}]\) and \([nv_0-\frac{eV_0}{h}]\)

(4)  \([n\frac{hv_0}{e}]\) and \([nv_1-\frac{eV_0}{h}]\)

Answer 1

 (2) \([(n-1)\frac{hv_0}{e}]\) and \([nv_0-\frac{eV_0}{h}]\)

From the graph for metal A, we see that its threhold frequency equals v₀ . Hence when light of frequency nv₀ falls on it, the maximum energy, of the emitted electrons, would equal (h (n -1) v₀). The corresponding magnitude, of the stopping potential would be \([\frac{(n-1)hv_0}{e}]\).

For metal B, the stopping potential being V₀ , the work function of this metal would equal (n.hv₀ - eV₀).

This implies that the threshold frequency, for this metal would be \(\frac{(nhv_0-eV_0)}{h}=(nv_0 -\frac{eV_0}{h}).\)