When certain metal surface is illuminated with a light of wavelength λ , the stopping potential is V, when the same surface is illuminated by light of wavelength 2 λ , the stopping potential is V 3 . The threshold wavelength for the surface is

Step-by-step solution

Given for a metal, wavelength of light used = λ stopping potential =V If λ 0 be the threshold wavelength, then maximum kinetic energy of emitted electrons K m a x = h c 1 λ - 1 λ 0 …(i) Again, wavelength of used light λ ' = 2 λ Stopping potential, V’ = V 3 then K m a x = h c 1 λ , 1 λ 0 ⇒ e V ' h c 1 2 λ - 1 λ 0 ⇒ e V 3 = h c 1 2 λ , 1 λ 0 …. (ii) From Eqs. (i) and (ii), we have e V e V 3 = h c 1 λ - 1 λ 0 h c 1 2 λ , 1 λ 0 3 1 2 λ - 1 λ 0 = 1 λ - 1 λ 0 ⇒ λ 0 = 4 λ So, threshold wavelength is 4 times of wavelength