When a metallic surface is illuminated by a light of frequency 8×10^14 Hz, photoelectron of maximum energy …..

Q. When a metallic surface is illuminated by a light of frequency 8×10¹⁴ Hz, photoelectron of maximum energy 0.5eV is emitted. When the same surface is illuminated by light of frequency 12 × 10¹⁴ Hz, photoelectron of maximum energy 2eV is emitted. The work function is

(a) 0.5 eV

(b) 2.85 eV

(c) 2.5 eV

(d) 3.5 eV

Ans: (c)

Sol: In first case ,

h(8×10¹⁴) = φ+0.5    ….(i)

In 2nd case ,

h(12 × 10¹⁴) = φ+2  ….(ii)

Dividing (ii) by (i)

h(12 × 10¹⁴) /h(8×10¹⁴)= (φ+2)/(φ+0.5)

3/2 =  (φ+2)/(φ+0.5)

2φ + 4 = 3φ + 1.5

φ = 2.5 eV