Q. Photons of energy 2.0 eV fall on a metal plate and release photoelectrons with a maximum velocity v . By decreasing λ by 25% the maximum velocity of photoelectrons is doubled. The work function of the metal of the material plate in eV is nearly
(a) 2.22
(b) 1.985
(c) 2.35
(d) 1.80
Ans: (d)
Sol: A/c to question , hc/λ = 2eV
In First case ,
hc/λ = φ + ½ (mv2)
hc/λ − φ = ½ (mv2) ……(i)
In 2nd case ,
hc/(3λ/4) = φ + ½ (m(2v)2)
hc/(3λ/4) − φ = ½ (m(2v)2) ….(ii)
Dividing (i) by (ii)
(hc/λ − φ)/(4hc/3λ − φ) = 1/4
4hc/λ − 4φ = 4hc/3λ − φ
4hc/λ (1 −1/3) = 3 φ
4hc/λ(2/3) = 3 φ
4×2×2/3 = 3 φ
φ = 16/9 =1.8 eV