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Question: If frequency (\(\upsilon > {\upsilon _0}\)) of the incident light becomes \(n\)times the initial fre...

If frequency (υ>υ0\upsilon > {\upsilon _0}) of the incident light becomes nntimes the initial frequency (υ\upsilon ) , the kinetic energy of the emitted photoelectrons becomes (υ0{\upsilon _0}is the threshold frequency)
A)nntimes of the initial kinetic energy.
B)more than nntimes of the initial kinetic energy.
C)Less than nntimes of the initial kinetic energy.
D)kinetic energy of the emitted photoelectron will remain unchanged.

Explanation

Solution

Hint: Einstein’s photoelectric equation tells us about the relation between the frequency of light used for emission of photoelectrons and the maximum kinetic energy of photoelectrons. Use this equation to deduce the expression.

Complete step by step solution:
Let the initial kinetic energy of the photoelectrons be KKand the new kinetic energy of the photoelectrons be K′K'.
For the first case, it is given that the frequency of incident light is υ\upsilon . It is also given in the question that this frequency is greater than the threshold frequency of the surface which means that the photoelectric effect will take place. The threshold frequency is υ0{\upsilon _0}.
C states that,
K=hυ−hυ0K = h\upsilon - h{\upsilon _0}
Where hhis the Planks’ constant.
From this equation we can find the value of hυh\upsilon
hυ=K+hυ0h\upsilon = K + h{\upsilon _0}
Now, the frequency is changed to nυn\upsilon
Again,
By Einstein’s photoelectric equation,
K′=nhυ−hυ0K' = nh\upsilon - h{\upsilon _0}
Putting the value of hυh\upsilon in this equation, we get,
K′=n(K+hυ0)−hυ0K' = n\left( {K + h{\upsilon _0}} \right) - h{\upsilon _0}
K′=nK+nhυ0−hυ0K' = nK + nh{\upsilon _0} - h{\upsilon _0}
K′=nK+hυ0(n−1)K' = nK + h{\upsilon _0}\left( {n - 1} \right)
From this equation, we can clearly see that the final kinetic energy of photoelectrons is greater than nn times of initial kinetic energy.
B) is correct.

Note: From Einstein’s photoelectric equation, students generally think that the kinetic energy of photoelectrons is directly proportional to the frequency of the incident radiation which is completely wrong. This is because the frequency term in the equation is in summation with one more term(work function).