Question

Consider the spectral line resulting from the transition n =2 to n=1 in the atoms and ions given below. The shortest wavelength is produced by:
A. Hydrogen atom
B. Deuterium atom
C. Singly ionised helium
D. Doubly ionised lithium

Answer 1

Hint: Use the formula for the wavelength of the electromagnetic radiation emit when the electron jumps from a higher energy level to a lower one. Compare the atomic numbers of the given atoms and ions to find the atom that produces the shortest wavelength.

Formula used:
$\dfrac{1}{\lambda }=R{{Z}^{2}}\left( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}} \right)$

Complete answer:
The wavelength of the emitted electromagnetic radiation when the electron undergoes transition from energy level ${{n}_{f}}$ and ${{n}_{i}}$ is given as:
$\dfrac{1}{\lambda }=R{{Z}^{2}}\left( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}} \right)$ …… (i).
Here, R is called Rydberg constant and Z is the atomic number of the atom or ion.
In the given case, in all the atoms and ions the transition from n = 2 to n = 1. Therefore, the value of ${{n}_{f}}$ and ${{n}_{i}}$ are the same for all four. Also R is a constant.
Therefore, the value of the wavelengths of the emissions in the four cases can be compared only with the value of Z and from equation (i) we get that the wavelength of the emission is inversely proportional to the atomic number of the atom or ion.
i.e. $\dfrac{1}{\lambda }\propto {{Z}^{2}}$.
This means that more the atomic number, lesser will be the wavelength.
Therefore, the shortest wavelength of will be produced by that atom or ion that is the greater atomic number amongst the four.
For a hydrogen atom, Z=1.
For a deuterium atom, Z=2.
For a singly ionised helium, Z=3.
For a doubly ionized lithium, Z=4
Hence, a doubly ionised lithium will produce the shortest wavelength.

This means the correct option is D.

Note:
Note that the formula for the wavelength of the emitted radiation in equation (i) is applicable only to species that have only one electron revolving around the nucleus.
That is why we discuss helium and lithium ions because they have a single electron. The atoms of these ions have multiple electrons and therefore, they cannot be considered.