When 1.00 mole of hydrogen atoms is excited from the n= 2 state to a single state of higher energy, 275.86 kJ

Question

What is the value of n for the higher state?

Can you also answer:
Calculate the uncertainty (in nm) in the position of an electron with the velocity of 1.50x103 m/s (the velocity is the same as in the preceding problem) and the uncertainty in the velocity is 1%. (2 significant figures are plenty)

Answer 1

What is the question? Perhaps find the n quantum number of the higher energy shell?. Specify.

All right.

The number of atoms in 1 mol is Na = 6.023x10^23 (Avogadro's number). So we need:

275.86x10^3/6.023x10^23 = 4.58x10^(-19) J of energy for one atom.

Suppose that n is the principal quantum number of the higher energy shell. Then

En - E2 = 4.58x10^(-19)

-2.18^10^(-18)/n^2 - (-2.18x10^(-18)/2^2) = 4.58x10^(-19)

where E1 = -2.18^10^(-18) J is the energy of the electron of the hydrogen atom in the shell with n = 1.

Solving the above equation for n you get:

n = 5

Answer 2

Rydberg consistent = a million.097373 X 10^7 according to meter. For convenience convert it into 'microns' (10^-6 meters), so as that Rydberg consistent (R) = 10.97373 / micron. The formula for use is a million/wavelength = R {(a million/n1^2) - (a million/n2^2)}. The preliminary state is given as n1 = 2. Longest wavelength is while n2 has a tendency to infinity. subsequently Wave length (longest) =a million/ R{(a million/4) - (0)} = 4 /10.97373 micron = 0.3645 micron This falls in 'seen' spectrum ('violet' end).