Question

In which of the following transitions will the wavelength be minimum?

(a) n = 5 to n = 4
(b) n = 4 to n = 3
(c) n = 3 to n = 2
(d) n = 2 to n = 1

Answer 1

(d) n = 2 to n = 1

For the transition in the hydrogen-like atom, the wavelength of the emitted radiation is calculated by

$\frac{1}{\lambda }=\mathrm{R}{Z}^2\left(\frac{1}{n_1}-\frac{1}{n_2}\right)$
Here, R is the Rydberg constant.

For the transition from n = 5 to n = 4, the wavelength is given by
$$\begin{gathered} \frac{1}{\lambda }=\mathrm{R}{Z}^2\left(\frac{1}{4^2}-\frac{1}{5^2}\right) \\ \lambda =\frac{400}{9\mathrm{R}{Z}^2} \end{gathered}$$


For the transition from n = 4 to n = 3, the wavelength is given by
$$\begin{gathered} \frac{1}{\lambda }=\mathrm{R}{Z}^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right) \\ \lambda =\frac{144}{7\mathrm{R}{Z}^2} \end{gathered}$$


For the transition from n = 3 to n = 2, the wavelength is given by
$$\begin{gathered} \frac{1}{\lambda }=\mathrm{R}{Z}^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ \lambda =\frac{36}{5\mathrm{R}{Z}^2} \end{gathered}$$


For the transition from n = 2 to n = 1, the wavelength is given by
$$\begin{gathered} \frac{1}{\lambda }=\mathrm{R}{Z}^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \\ \lambda =\frac{2}{\mathrm{R}{Z}^2} \end{gathered}$$

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.