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Radiation of frequency 1015 Hz is …

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Radiation of frequency 1015 Hz is incident on three photosensitive surfaces A, B and C. Following observations are recorded: Surface A: no photoemission occurs Surface B: photoemission occurs but the photoelectrons have zero kinetic energy. Surface C: photo emission occurs and photoelectrons have some kinetic energy. Using Einstein’s photo-electric equation, explain the three observations

Posted by Jatin Grewal 2 years, 2 months ago

CBSE > Class 12 > Physics

1 answers

Anjan Karthi 2 years, 2 months ago

A must have had a work function greater than 1015h. So, no matter how intense the light is or for how long the radiation has been falling on the surface, neither would electrons gain energy to be emitted out of the surface. But, in the case of B (as per Einstein equation " hv = hv° + eV° "), emission happened but eV° is ZERO. So, Work function = 1015h for B. For C, Work function would be much lesser than 1015h due to which the surface could emit electrons with some eV° with radiation of 1015 Hz.

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