A photoelectric surface is illuminated successively by monochromatic light of wavelength $λ and \frac{λ}{2} .$ 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)

Question

A photoelectric surface is illuminated successively by monochromatic light of wavelength $λ and \frac{λ}{2} .$ 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)

Infinity Learn · Accepted Answer

Let ϕ$$0$$  be the work function of the surface of the material. Then, According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons in the first case $$isKmax1=hc$$λ$$-$$ϕ$$0and$$ that in the second case $$isKmax2=hc$$λ$$2-$$ϕ$$0=2hc$$λ$$-$$ϕ$$0$$ But $$Kmax2=3Kmax1($$ given $$)$$∴  $$2hc$$λ$$-$$ϕ$$0=3hc$$λ$$-$$ϕ$$02hc$$λ$$-$$ϕ$$0=3hc$$λ$$-3$$ϕ$$03$$ϕ$$0-$$ϕ$$0=3hc$$λ$$-2hc$$λ$$2$$ϕ$$0=hc$$λ   or   ϕ$$0=hc2$$λ

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