Physics-Electrons,Waves and Photons AS Level June 2011

Revising resisitance, resistivity and waves. so difficult to find revision. anyone got any good notes or know any good sites?

Yep it does thank you!

But then if the actual y value is not the photon energy (because the photon energy = the y value + energy needed to release the electron) then how can we just use the y-value/frequency to work out Planks constant, should it not be the (y-value + energy needed to release electron)/frequency?

anyone get jan 10 question 7bii and 8ci

hey,

cud anyone explain to me the term stopping potential?

thanks.

The photon theory that explains the photoelectric effect proposes that it is the interaction of individual photons with individual electrons; that is, each interaction involves one photon and one electron. Einstein's Photoelectric Equation $hf = \phi + KE_{MAX}$ describes this; it is a way of looking at the interaction between one photon and one electron in terms of the conservation of energy. The incoming photon has energy $hf$. Some of this is needed to overcome the work function $\phi$ of the metal; the rest is transferred to the kinetic energy of the electron.

So, onto your actual question... Increased intensity just meand more photons (since they all have the same frequency). The above equation shows that the maximum kinetic energy of any emitted electron is $hf - \phi$; there may be more of them if you increase the intesity (or fewer of them if you decrease the intesity), but the maximum kinetic energy of each emitted electron only depends on the frequency of the light and the work function of the metal.

However, the fact that increased intensity means more photons (again, assuming the frequency of the light is unchanged) means that increased intensity leads to more emitted electrons, and therefore a larger current in a photocell.

Do you appreciate why the above equation includes a term for the maximum kinetic energy of an emitted electron, rather than simply the kinetic energy of such an electron?

As it is the max Ke needed to emit an electron from the surface of the metal

This is still not quite right. You are correct to say that the photon gives up all of its energy to the electron (although it is best to say that it is absorbed by the electron rather than it hits the metal; the latter suggests that the photon may still exist after the interaction, which it does not). So, of the initial $hf$ photon energy, $\phi$ goes to overcoming the work function of the metal. This leaves $hf - \phi$. This is exactly the kinetic energy of the electron immediately after the interaction - conservation of energy will not allow anything else. But before the electron can leave the metal, it may collide with other atoms in the metal, losing kinetic energy on the way. So, by the time it emerges from the metal, it may have less kinetic energy than it could have had, but it could never have more than $hf - \phi$. This is its maximum kinetic energy, $KE_{MAX}$.

Einstein's Photoelectric Equation describes a conservation analysis of a single photon-electron interaction. But any individual ejected electron may have a kinetic energy of anything up to the permitted maximum.

Hey does anyone how to work out the answers for Q4 e and f from JANUARY 2011?

4e revolves around the relationship between the intensity of a wave and its amplitude. Do you know what that is?

In 4f, you know that the two waves are initially in antiphase. You can also deduce from the graphs that they have the same wavelength (how?). Does that tell you how far one source has to be moved to bring them into phase?

Yh, for e.i, i know u have to use the fact that intensity is proportional to A^2 , BUT THEN I DONT get the one below it ii.