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Photoelectric emission question

In a demonstration, ultraviolet light is incident on a zing plate and electrons are emitted.
The intensity of the ultraviolet light is increased.
Explain the following observations:
- the number of electrons emitted per second increases
- the maximum kinetic energy of an electron does not change
My answer so far:
Because more photons are being emitted per second, thus higher probability of energy of the photon being transferred to a single electron.
How do I prove that the maximum energy of an electron does not change?
I know I have to use

hf = \Phi+\frac{1}{2}mv^2

and

\Phi

is a constant which is specific to the metal and the electrons mass doesn't change.

Reply 1

Dynamo123

Constant frequency, constant max.K.E energy.
The first answer is correct.

Reply 2

halpme

Original post by Dynamo123

Constant frequency, constant max.K.E energy.
The first answer is correct.

Does that mean that

\frac{1}{2}mv^2

remains constant for any given work function (

\Phi

\Phi

is the only thing that can change in that equation, right? Along with

v

Reply 3

Thr33

The energy is affected by the work function (which is constant for a given piece of metal), but, more importantly, the frequency of the light. (hf and all that). If you're calculating the KE, you don't need to worry about the velocity of the particles...(as it's directly proportional to the KE you've just found).

Reply 4

krisshP

Original post by halpme

Does that mean that

\frac{1}{2}mv^2

remains constant for any given work function (

\Phi

\Phi

is the only thing that can change in that equation, right? Along with

v

\Phi

depends on the material, so zinc has certain work function while another material would have a different work function, but it's constant as only zinc's used.

This is what I'd say

One photon releases one electron.
Increasing intensity only increases the number of electrons emitted per second, not the light frequency.
By

hf=\Phi + \frac{1}{2} mv^2

Frequency affects the maximum kinetic energy of the emitted electrons, but intensity doesn't.

(edited 10 years ago)

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