Atoms — Physics STD 12 Science — Question

For the transition in the hydrogen-like atom, the wavelength of the emitted radiation is calculated by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{\text{n}_1}-\frac{1}{\text{n}_2}\Big)$
Here, $R$ is the Rydberg constant.
For the transition from $n = 5$ to $n = 4$, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{4^2}-\frac{1}{5^2}\Big)$
$\lambda=\frac{400}{9\text{RZ}^2}$
For the transition from $n = 4$ to $n = 3$, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{3^2}-\frac{1}{4^2}\Big)$
$\lambda=\frac{144}{7\text{RZ}^2}$
For the transition from $n = 3$ to $n = 2$, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{2^2}-\frac{1}{3^2}\Big)$
$\lambda=\frac{36}{5\text{RZ}^2}$
For the transition form $n = 2$ to $n = 1$, the wavelength is given by,
$\frac{1}{\lambda}=\text{RZ}^2\Big(\frac{1}{1^2}-\frac{1}{2^2}\Big)$
$\lambda=\frac{2}{\text{RZ}^2}$
From the above calculations, it can be observed that the wavelength of the radiation emitted for the transition from $n = 2$ to $n = 1$ will be minimum.