Question

Consider the spectral line resulting from the transition from $n=2$ to $n=1$, in atoms and ions given below. The shortest wavelength is produced by:

• A. hydrogen atom
• B. deuterium atom
• C. singly ionized helium
• D. doubly ionized lithium

Answer 1

The correct option is D doubly ionized lithium

Wavelength is calculated as

$\frac{1}{\lambda }=R{z}^2(\frac{1}{n_f^2}=\frac{1}{n_i^2})$

$\therefore \lambda \propto \frac{1}{z^2}$

For hydrogen atom , z=1

For Deuterium atom , z=2

For singly ionized helium, z=2

For doubly ionized lithium, z=3

Thus, z is maximum for lithium and so $\lambda$ is minimum.