00:01
For this problem on the topic of quantum mechanics, we have an electron in a cubicle box of side length l, initially in the ground state.
00:09
The longest wavelength photon that is absorbed has a wavelength in a of 624 nanometers, and we want to find the side length l.
00:17
The wavelength lambda is found to be 234 nanometers, which is also absorbed when the initial state is still in the ground state.
00:25
We want to then find the value of n squared for the final state in the transition for which this wavelength is.
00:30
Absorbed where we know n squared is nx squared plus n y squared plus n z squared we then want to find the degeneracy of this energy level now the energy levels of a particle in a cubicle box of length l are e n x n x n y and z and this is equal to n x squared plus n y squared plus n z squared times pi squared square over 2ml squared.
01:07
And the longest wavelength absorbed is for the smallest energy transition, which is between the ground state 111 and the next highest state, 211.
01:16
The energy absorbed by the photon, delta e is hc over lambda.
01:21
So for the lowest energy transition, we have delta e equal to 2 squared plus 1 squared, plus 1 squared minus 1 squared plus 1 squared plus 1 squared into pi squared h bar squared over 2m l squared and this is equal to the energy of the photon delta e which is hc over lambda or hc over 624 nanometers and so from here we get by rearranging that the length, beside length of the box, l is equal to the square root of three times times constant h times 624 nanometers divided by 8mc.
02:29
Putting in our values, we get this to be the square root of 3 into 6 .6262.
02:39
6 times 10 to the minus 34 joules seconds, which is blanks constant, times 624 nanometers, divided by 8 times 9 .11 times 10 to the minus 31 kg, the mass of the electron, times the speed of light and vacuum, 3 times 10 to the power 8 meters per second...