Q: The work function of a metal is 3.0 eV. It is illuminated by a light of wave length 3 × 107 m. Calculate (i) threshold frequency, (ii) the maximum energy of Photoelectrons , (iii) the stopping potential. (h = 6.63 × 10-34) Js and c = 3 × 108 ms-1).
Solution: (i) W = h ν0 = 3.0 eV = 3 x 1.6 x 10^(-19) J
Threshold frequency ,
ν0 = W/h = (3 × 1.6 × 10-19)/(6.63 × 10-34 )
= 0.72 × 1015 Hz.
(ii) Maximum kinetic energy , Kmax = h (v – v0)
λ = 3 × 10-7m,
ν = c/λ = (3 × 108)/(3 × 10-7 )
= 1 × 1015 Hz
Kmax = h (ν – ν0)
Kmax = 6.63 × 10^(-34) (1-0.72) × 10^15 J
= 1.86 × 10-19 J.
(iii) Kmax = e V0 , where V_0 is stopping potential in volt and e is the charge of electron
V0 = Kmax/e . Here, Kmax = 1.86 × 10-19 J
e = 1.6 × 10-19 C;
V0 = (1.86 × 10-19)/(1.6 × 10-19 ) = 1.16 V