1.

A photon of wave length 4 x 10-7m strikes on metal surface, work function of the metal being 2.13 eV.Calculate –(i) The energy of the photon (eV) (ii) The kinetic energy of the emission (1 eV= 1.6020 × 10-19J)

Answer»

(i) Energy (E) = \(\frac{hv}{\lambda}\); h = 6.6 × 10−34Js

C = 3 x 108 ms-1

E = \(\frac{6.6\times10^{-34}Js\times3\times10^8m}{4\times10^{-7}m}\)

= 4.97 × 10−19 J

\(\frac{4.97\times10^{-19}J}{1.602\times10^{-9}}\)

= 3.10 eV.

Energy of the photon in eV = 3.10

(ii) Kinetic energy of the emission of photo-electrons (K.E)

= hv − \(\omega_o\)(where \(\omega_o\) = work function)

= 3.10 eV − 2.13 eV = 0.97 eV.



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