Question

For a constant frequency of the incident light, the graph between frequency (ν) of the incident light, and the magnitude (Vs), 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

Answer 1

Correct option : (2)

Explanation :

From the graph for metal A, we see that its threshold 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

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