1. When the electromagnetic radiations of frequencies 4*10^15 Hz and 6*10^15 HZ fall on the same metal in different experiments, the ratio of maximum kinetic energy of electrons liberated is 1 : 3. the threshold frequency for metal is

For the first case, h = planck 's constant, c = speed of light,

ν1 =first frequency

ν2= second frequency

νo = threshold frequency
As kinetic energy is given by :-
K.E. = ν- νo h   

h(ν1 -νo)/h(ν2-νo) = 1/3

=> 3ν1 -3νoν2νo

=> 2νo = 3ν1 - ν2

=> νo = (3ν1ν2)/2 = (12 - 6) * 1015 /2 = 3 x 1015Hz.

 

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