When radiation of wavelength λ is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4 . 8 V . If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1 . 6 V . The threshold wavelength of the metal is:

Step-by-step solution

V S = h v - ϕ 4 . 8 = h c λ - ϕ     . . . ( i ) 1 . 6 = h c 2 λ - ϕ   . . . ( ii ) Using above equation ( i ) - ( ii ) 3 . 2 = h c λ - h c 2 λ 3 . 2 = hc 2 λ       . . . ( iii ) λ = h c 6 . 4 Put in equation ( ii ) ϕ = 1 . 6 h c λ th = 1 . 6 λ th = h c 1 . 6 = h c 6 . 4 × 4 = 4 λ