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+1 vote
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in Dual Nature of Radiation and Matter by (48.5k points)
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Light of wavelength 5000 Å falls on a metal surface of work function 1.9 eV. 

Find: 

(i) the energy of photons in eV 

(ii) the K.E of photoelectrons and 

(iii) the stopping potential.

1 Answer

+1 vote
by (48.1k points)
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Best answer

Here λ = 5000 Å = 5 x 10-7

W0 = 1.9 ev

(i) Energy of a photon,

E = 2.475 eV

(ii) K.E of a photoelectron, 

K.E = hυ – W0 = 2.475 – 1.9 = 0.575 eV 

(iii) Let V0 be the stopping potential. Then 

eV0\(\frac{1}{2}\)mv2 = K.E of a photoelectron

 V0\(\frac{0.575}{e}\) eV 

V0 = 0.575 V

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