Question

considering the transitions in a hydrogen atom, which transition results in the emission of a photon with the shortest wavelength?
n=5 ----> n=1
n=4 ----> n=1
n=2 ----> n=5
n=3 ----> n=5
n=1 ----> n=4

Explanation:

Step1: Recall energy - wavelength relation

The energy of a photon is given by $E = h
u=\frac{hc}{\lambda}$, where $h$ is Planck's constant, $
u$ is frequency, $c$ is the speed of light and $\lambda$ is wavelength. Shorter wavelength means higher - energy photon.

Step2: Recall hydrogen - atom energy levels

For a hydrogen atom, the energy of an electron in the $n^{th}$ level is given by $E_n=-\frac{13.6}{n^{2}}\text{ eV}$. When an electron transitions from a higher level $n_i$ to a lower level $n_f$, the energy of the emitted photon is $\Delta E = E_{n_i}-E_{n_f}=13.6(\frac{1}{n_f^{2}}-\frac{1}{n_i^{2}})\text{ eV}$.

Step3: Calculate energy for each transition

For $n = 5

ightarrow n = 1$:
$\Delta E_1=13.6(\frac{1}{1^{2}}-\frac{1}{5^{2}})=13.6(1 - \frac{1}{25})=13.6\times\frac{24}{25}=13.056\text{ eV}$

For $n = 4

ightarrow n = 1$:
$\Delta E_2=13.6(\frac{1}{1^{2}}-\frac{1}{4^{2}})=13.6(1-\frac{1}{16})=13.6\times\frac{15}{16}=12.75\text{ eV}$

For $n = 2

ightarrow n = 5$:
This is an absorption transition (not emission as we need), since the electron is going from a lower to a higher energy level.

For $n = 3

ightarrow n = 5$:
This is also an absorption transition.

For $n = 1

ightarrow n = 4$:
This is an absorption transition.

Step4: Compare energies

The transition with the highest - energy photon (and thus the shortest wavelength) is the one with the largest energy difference. The $n = 5
ightarrow n = 1$ transition has the largest energy difference among the emission transitions considered.

Answer:

$n = 5
ightarrow n = 1$