- 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,
1 =first frequency
2= second frequency
o = threshold frequency
As kinetic energy is given by :-
K.E. = h
h(1 -o)/h(2-o) = 1/3
=> 31 -3o = 2 -
=> 2o = 31 - 2
=> o = (31 - 2)/2 = (12 - 6) * 1015 /2 = 3 x 1015Hz.
1 =first frequency
2= second frequency
o = threshold frequency
As kinetic energy is given by :-
K.E. = h
h(1 -o)/h(2-o) = 1/3
=> 31 -3o = 2 -
=> 2o = 31 - 2
=> o = (31 - 2)/2 = (12 - 6) * 1015 /2 = 3 x 1015Hz.