Atomic Structure 1 Question 27

3. What is the work function of the metal, if the light of wavelength 4000\AA generates photoelectron of velocity 6×105 ms1 from it?

(Mass of electron =9×1031 kg

Velocity of light =3×108 ms1

Planck’s constant =6.626×1034Js

Charge of electron =1.6×1019JeV1 )

(2019 Main, 12 Jan I)

(a) 4.0eV

(b) 2.1eV

(c) 0.9eV

(d) 3.1eV

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Answer:

Correct Answer: 3. (d)

Solution:

  1. Work function of metal (ϕ)=hv0 where, v0= threshold frequency

Also, 12mev2=hvhv0

or

12mev2=hvϕ 12mev2=hcλϕ

Given : λ=4000\AA=4000×1010 m

v=6×105 ms1, me=9×1031 kg,c=3×108 ms1 h=6.626×1034Js

Thus, on substituting all the given values in Eq. (i), we get

12×9×1031 kg×(6×105 ms1)2 =6.626×1034 J s×3×108 ms14000×1010 mϕ ϕ=1.62×1021kgm2 s24.96×1019 J =3.36×1019 J[1 kg m2 s2=1 J] =2.1eV