Kinetic energy of the emitted electrons

[xiphoid]

Homework Statement

When the frequency of light incident on a metallic plate is doubled, the K.E. of the emitted photoelectrons will be?

Homework Equations

The answer should be either doubled or increased but more than double of the previous kinetic energy

The Attempt at a Solution

λ= h/ √2mK.E.

therefore, 1/v= h/c*√2mK.E.

 

[AGNuke]

K.E. of photoelectron = (KE of incident photon) - (Work Function of Metal).

Since you doubled the energy of incident photon, it will no doubt increase the KE of photoelectron, but the extra energy will be purely dedicated in doing so, unlike before doubling the energy, in which some of the energy was required to free electron from the metal.

So, the new KE will be more than double the initial KE.

 

[dharavsolanki]

K.E. of photoelectron = (KE of incident photon) - (Work Function of Metal).

Let's use this equation and say that this is the starting point.

Since you doubled the energy of incident photon, it will no doubt increase the KE of photoelectron

Now, you have doubled the incident photon's energy, which is one term on the right side. However, the second term, the work function, is still the same. This is the crux of the problem.

K.E. of photoelectron 1 = (KE of incident photon 1) - (Work Function of Metal).

K.E. of photoelectron 2 = (KE of incident photon 2) - (Work Function of Metal).

Now, rewriting this energy in terms of the energy of incident photon,

K.E. of photoelectron 2 = 2*(KE of incident photon 1) - (Work Function of Metal).

Difference of the two energies = KE of incident photon 1

Question: How did I get this relation? Simple enough, but just saves me some typing

So, the energy of the second photoelectron EXCEEDS that of the earlier one by a magnitude equal to the KE of incident photon 1.

BUT, KE of incident Photon 1 is MORE than the KE of photoelectron 1 (by first equation.)

So, yes, answer is more than twice of the initial photoelectron.

Intuition

Think of the problem this way - You have to hire a plumber. You have to pay for the services and his transportation cost. BUT, if your neighbor decides to have the plumbing done too, the two of you will not pay the transportation cost twice.

So, effectively, the cost per person is LESS than that if only one person would have asked for plumber.

It's really a fixed cost, variable cost problem which we learned in our linear equation days.

Likewise, the work function has to be paid off only once, for the electron to surface. It has got nothing to do with the incident energy of the photon.