## Kinetic Energy

Bao Tram Nguyen
Posts: 35
Joined: Sat Aug 24, 2019 12:17 am

### Kinetic Energy

What happens to the kinetic energy in Ek=(1/2)Me-Ve-^2 if we were to increase the frequency or intensity of the electrons? Or when does the kinetic energy change, if it changes at all.

William Francis 2E
Posts: 101
Joined: Wed Sep 18, 2019 12:20 am

### Re: Kinetic Energy

Frequency and intensity are variables manipulated for the light shining on the metal surface in the photoelectric experiment rather than for the electrons. In the equation Ek=1/2(me-)(ve-)^2, the "m" variable represents the mass of the electron (which I believe should be a constant), and the "v" variable represents the velocity of the electron. Electrons emitted with high velocity have more kinetic energy, and those emitted with lower velocity have lower kinetic energy.

Trinity Vu 1D
Posts: 53
Joined: Fri Aug 30, 2019 12:15 am

### Re: Kinetic Energy

Adding on, increasing the intensity of the light shined on the metal doesn't affect the kinetic energy because the kinetic energy is in regards to the kinetic energy per electron released not total kinetic energy. Increasing the intensity of the light would mean increasing the number of photons and thus increasing the number of electrons ejected. However, the kinetic energy of each electron would remain the same thus resulting in no change of Ek.

Chetas Holagunda 3H
Posts: 51
Joined: Thu Jul 11, 2019 12:17 am
Been upvoted: 1 time

### Re: Kinetic Energy

I would think that as you increase the intensity or frequency, more energy would be given off by the photon, surpassing the threshold energy. Since the energy needs to be conserved, the velocity of the electron would be greater as the energy given by the photon increases since kinetic energy's variable value is its velocity while mass of the electron stays the same.

Shutong Hou_1F
Posts: 117
Joined: Sat Sep 14, 2019 12:17 am

### Re: Kinetic Energy

Yes, the kinetic energy does increase if we were to increase the frequency.
In solving photoelectric effect problem, we generally don't directly use this equation (Ek = (1/2) Me- * Ve-^2) to get the kinetic energy, probably because the Ve is hard to directly measure and thus unknown to us in problem and also in experiments.
Instead, we usually use E(photon) - E(energy remove e-) = Ek(e-), h * frequency - work function = Ek.