Question

Which of the following transitions (in a hydrogen atom) represent emission of the longest wavelength photon?
a) n = 1 to n = 2
b) n = 3 to n = 1
c) n = 3 to n = 4
d) n = 4 to n = 2
e) n = 5 to n = 4

Answer 1

For a photon to be emitted, the electron must travel from a higher quantum level to a lower one. Therefore, options A and C are eliminated.
The energy of a photon is given by Planck's equation:
E = hc/λ; where λ is the wavelength
As seen from the equation, a larger wavelength is observed when smaller energy transitions occur (since energy and wavelength are inversely proportional)
The smallest transition occurs in option E, 5 to 4. Therefore, the answer is E.

Answer 2

Answer:

E) n = 5 to n = 4

Explanation:

Since the question stated that it is an emission process, options involving increase in transitions are to be ignored. This means option a and b are to be ignored as they refer to absorption.

We are left with three options.

The second option can be dropped because of the huge energy gap involved. Any drop into level one is in the ultraviolet, which are short wavelengths.

We are left with two options now.

The fouth is a transition between two levels, ehich means extra energy involved.

The fifth is a transition between just one level, which means lesser energy is involed.

Energy and wavelength are inversely proportional to each other, so the higher the energy, the lesser the wavelenth.

The emission with the longest wavelength (or, least energy) is the last option, the 5 to 4 transition.