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### Photoelectric effect

Posted: Sun Oct 04, 2015 10:16 pm
Hey guys so I was wondering if you could help me understand why short wavelength light can eject electrons from a metal surface, while longer wavelengths don't? Still kind of confused on this concept.

### Re: Photoelectric effect

Posted: Sun Oct 04, 2015 10:28 pm
short wavelengths have more energy, i.e. it has enough energy to "shake" the electron loose. alternatively, a longer wavelength carry less energy. at a certain point, the wavelength becomes too long (not enough energy) to "shake" the electron loose.

### Re: Photoelectric effect

Posted: Mon Oct 05, 2015 1:35 am
A photon has to meet a threshold energy to eject an electron. Based on the equation $c=\lambda \nu$, we can see that frequency and wavelength have an inverse relationship because c, the speed of light, is a constant. If the wavelength goes up, the frequency goes down. If the wavelength decreases, the frequency of light increases. We then look at the equation $E=h\nu$, showing that light with a higher frequency has more energy. The high frequency of high energy light also means that the wavelength is low due to the inverse relationship. Comparatively, high wavelength light would have a lower frequency and therefore less energy.

However, both high and low energy light can cause an electron to be ejected from a metal as long as the energy exceeds the energy threshold or work function of that particular metal.

### Re: Photoelectric effect

Posted: Mon Oct 05, 2015 2:12 pm
When the wavelength is shorter, the frequency of the wave increases which means that the wave has more energy. Since in this experiment light acts like particles, that means the particles have more energy and are therefore able to eject electrons.

### Re: Photoelectric effect

Posted: Tue Oct 06, 2015 10:31 am
Is ultraviolet light the longest wavelength that is effective in the photoelectric effect or is it just dependent on the type of metal being used?

### Re: Photoelectric effect

Posted: Tue Oct 06, 2015 12:19 pm
Whether the photoelectric effect occurs is dependent on the work function which is a property is the metal used.

### Re: Photoelectric effect

Posted: Sun Oct 11, 2015 10:06 pm
What is the difference between the equations E=(-hR/n^2) and v=R(1/n^2 - 1/n^2)? Can these equations be used interchangeably?

### Re: Photoelectric effect

Posted: Thu Oct 15, 2015 4:21 pm
Chem_Mod wrote:Whether the photoelectric effect occurs is dependent on the work function which is a property is the metal used.

Thank you! Also I had another question. If we were doing an uncertainty problem trying to find the uncertainty of the position of something traveling +/- 0.7 mxs^-1
do we have to double 0.7 to 1.4 in the calculations or do we split u the problem into a +0.7 and a -0.7 and end up having two answers. Or do we just use 0.7?

### Re: Photoelectric effect

Posted: Thu Oct 15, 2015 7:09 pm
204578727 wrote:
Chem_Mod wrote:Whether the photoelectric effect occurs is dependent on the work function which is a property is the metal used.

Thank you! Also I had another question. If we were doing an uncertainty problem trying to find the uncertainty of the position of something traveling +/- 0.7 mxs^-1
do we have to double 0.7 to 1.4 in the calculations or do we split u the problem into a +0.7 and a -0.7 and end up having two answers. Or do we just use 0.7?

The uncertainty would be 1.4.