Radiation of wavelength λ is incident on a photocell. The fastest emitted electron has to speed v. If the wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be:

$less than straight v space open parentheses 4 over 3 close parentheses to the power of 1 divided by 2 end exponent$

= v(4/3)^1/2

=v(3/4)^1/2

Solution

According to the law of conservation of energy, i.e. Energy of a photon (hv) = work function (Φ) + Kinetic energy of the photoelectron (mv²/2)
according to Einstein's photoelectric emission of light,
E =(KE)_max + Φ
As, hc/λ = (KE)_max + Φ
If the wavelength of radiation is changed to 3λ/4
then,
$4 over 3 hc over straight lambda space equals space open parentheses 4 over 3 left parenthesis KE right parenthesis subscript max space plus space straight capital phi over 3 close parentheses plus straight capital phi For space fastest space emitted space electron comma space left parenthesis KE right parenthesis subscript max space equals space 1 half space mv squared space plus straight capital phi fraction numerator begin display style 1 end style over denominator begin display style 2 end style end fraction space mv squared space space equals space 4 over 3 space open parentheses 1 half mv squared close parentheses space plus straight capital phi over 3 straight i. straight e comma space straight v apostrophe greater than space straight v open parentheses 4 over 3 close parentheses to the power of 1 divided by 2 end exponent$

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