Question

A photoelectric surface is illuminated successively by monochromatic light of wavelength λ and λ/2. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is:(h = Planck's constant, c = speed of light)(a) $\frac{hc}{\lambda}$(b) $\frac{2hc}{\lambda}$(c) $\frac{hc}{3\lambda}$(d) $\frac{hc}{2\lambda}$

Answer 1

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Answer 2

Answer:

D) hc/2λ

Explanation:

Let ϕo be the work function of the surface of the material. Then, According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photo electrons in the first case will be -

E = Kmax + ϕ

where,

KE1 = hcλ−W (eq 1)

3K.E1 = 2hc/λ−W

= K.E1 =2hc/3λ−W/3 (eq 2)

= hc/λ−W =2hc/3λ−W/3

On equating (1) and (2) we will get

= hc/λ(1/3) = 2W/3

ϕo = W = hc/2λ

If the maximum kinetic energy of the emitted photo electrons in the second case is 3 times that in the first case, the work function of the surface of the material is hc/2λ