Answer:
d. The shape of the corresponding atomic orbital(s)
Explanation:
in the Hydrogen atom n gives us the energy of the electron that is orbitating the nucleus, given by.
[tex]E_{n} = -\frac{ K Z^{2} e^{4} }{n^{2}}[/tex], where K is a constant, Z is the number of protons of the nucleus, and e is the electron's charge.
-a. the option a must be true, given that that's exactly what the above equation it's for.
-b. n can have the values [tex]n={1,2,3,4 ...}[/tex] so if you replace n = 1 in the above equation then you will get the lowest (negative) energy of the electron, so for you to remove the electron you will need to give the biggest amount of energy compared with the other energy level's available for the electron, so that means that the maximum energy and therefore the maximum frecuency (minimum wavelength) needed to remove the electron and will be from the wave wich energy is equal to the energy calculeted using n = 1. So that means b is true.
-c, n is directly related to the orbital's size given that if you plot the graphs of the orbitals you will then realise that as you increase n then it is increased the size of the orbital, and viceversa for example (1s -> n=1, 2s->n=2, etc ...) i.e 1s is smaller than 2s , so c it's true.
-d, If you do the same excersise and plot the graph of the orbitals for the orbitals for example: 2px->n=2, 2py->n=2 and 2pz->n=2. They are quite different but have the same principal quantum number, that is n=2. So the shape of the orbital is'nt entirely given by the principal quantum number n. So therefore d is not true.
