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Question: When photons of energy \(h\nu \) fall on a metal plate of work function \('{{W}_{0}}'\), photoelectr...

When photons of energy hνh\nu fall on a metal plate of work function ′W0′'{{W}_{0}}', photoelectrons of maximum kinetic energy ′K′'K' are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the photoelectrons will be?
A) K+W0A)\text{ }K+{{W}_{0}}
B) K+hνB)\text{ }K+h\nu
C) KC)\text{ }K
D) 2KD)\text{ 2}K

Explanation

Solution

This problem can be easily solved using Einstein’s photoelectric equation which relates the maximum kinetic energy of the ejected photoelectron with the work function of the metal and the energy of the photon falling on it. By writing the equation for the doubled frequency and comparing it with the standard equation, we can get the required answer.

Formula used:
hν=K+W0h\nu =K+{{W}_{0}}

Complete step by step solution:
We will use Einstein’s photoelectric equation for the two cases and compare them to the standard equation to get the required result.
According to Einstein’s photoelectric equation, the maximum kinetic energy KK of the ejected photoelectron when a photon of energy hνh\nu falls on a metal surface with work function W0{{W}_{0}} is given by
hν=K+W0h\nu =K+{{W}_{0}} --(1)
Now let us analyze the question.
We are given that a photon of energy hνh\nu falls on a metal surface with work function W0{{W}_{0}} and the maximum kinetic energy of the photoelectron ejected is KK.
Therefore, using (1), we get,
hν=K+W0h\nu =K+{{W}_{0}} --(2)
Now, the frequency ν\nu is doubled.
Therefore, the energy of the photon will become h(2ν)=2hνh\left( 2\nu \right)=2h\nu
Let the maximum kinetic energy of the ejected photoelectron in this case be K′K'. The work function will still remain the same, that is, W0{{W}_{0}} since it is a property of the metal and not dependent upon the frequency of the radiation falling on it.
Therefore, using (1), we get,
2hν=K′+W02h\nu =K'+{{W}_{0}} --(3)
Plugging in (2) in (3), we get,
2(K+W0)=K′+W02\left( K+{{W}_{0}} \right)=K'+{{W}_{0}}
∴2K+2W0=K′+W0\therefore 2K+2{{W}_{0}}=K'+{{W}_{0}}
∴K′=2K+2W0−W0=2K+W0=K+K+W0\therefore K'=2K+2{{W}_{0}}-{{W}_{0}}=2K+{{W}_{0}}=K+K+{{W}_{0}}
∴K′=K+hν\therefore K'=K+h\nu [Using (2)]
Therefore, the maximum kinetic energy of the photoelectron as required by the question will be K+hνK+h\nu .

Hence, the correct option is B) K+hνB)\text{ }K+h\nu .

Note: Students should not think that if the frequency of the radiation is doubled, it will have an effect on the work function of the metal surface also. The work function of the metal surface is a constant characteristic property of the metal and is independent of the frequency of the radiation falling on it.
Students must also remember that Einstein’s photoelectric equation gives the maximum kinetic energy that an ejected photoelectron can have. All ejected photoelectrons do not have this kinetic energy. Most of them suffer collisions with other electrons and transfer part of their kinetic energy. The maximum kinetic energy is the case when the photon completely transfers all its energy to a single electron and it also does not suffer any other collisions.