When radiation of the wavelength is incident on a metallic surface, the stopping potential is 4.8 ~V . If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes 1.6 ~V . Then, the threshold wavelength for the surface is :

Step-by-step solution

Stopping potential. V₀= h c e ( 1 - 1 ₀ ) where, V₀= stopping potential Ist case, 4.8= h c e ( 1 - 1 ₀ ) IInd case, 1.6= h c e ( 1 2 - 1 ₀ ) Dividing Eqs. (i) by (ii), 3= ( 1 - 1 ₀ ) ( 1 2 - 1 ₀ ) 3 2 - 3 ₀ = 1 - 1 ₀ 1 ₀ - 3 ₀ = 1 - 3 2 -2 ₀ = 2-3 2 2 ₀ = 1 2 ₀=4