00:02
For this problem, the number of electrons that can be excited from the valence band to the conduction band, we will denote it as n.
00:10
This n should be equal to the total energy that is available to the electrons, e divided by the amount of energy that is needed to excite a single electron, e0.
00:24
So we will compute this e and e0 separately.
00:27
For e, this energy is provided by the gamma ray photon.
00:32
For a photon, we know the energy is h times f, where h is the plunk constant, and f is the frequency of the photon.
00:42
In this problem, we're not given the frequency, but we're given the wavelength.
00:46
So we will need to use the relationship.
00:48
F equals c over lambda, where c is the speed of light, and lambda is the wavelength.
00:55
So plug this in.
00:59
We get e equals h, c, the lambda.
01:02
And now we use the numbers.
01:06
The plug constant is 6 times 64 times 10 to the negative 34 dual second, and speed of light is 3 times 10 to the 8 in a per second.
01:24
And wavelengths are given the problem, which is 9 .31 times 10 to the negative 4 nanometer.
01:37
However, here we need to be careful...