Question

A hydrogen atom in state n = 6 makes two successive transitions and reaches the ground state. In the first transition a photon of 1.13 eV is emitted. (a) Find the energy of the photon emitted in the second transition (b) What is the value of n in the intermediate state?

Answer 1

Energy (E) of the nth state of hydrogen atom is given by
E = $\frac{13.6}{n^2}$ eV

For n = 6,

$\therefore E=\frac{-13.6}{36}=-0.377777777$ eV

Energy of hydrogen atom in the ground state = −13.6 eV
Energy emitted in the second transition = Energy of ground state $-$ (Energy of hydrogen atom in the 6th state + Energy of photon)
$$\begin{gathered} =13.6-\left(0.3777\overline{7}+1.13\right) \\ =12.09=12.1\mathrm{eV} \end{gathered}$$

(b)
Energy in the intermediate state = (Energy of photon emitted in the first transition) + (Energy of n = 6 state)
= 1.13 eV + 0.377 eV
= 1.507 eV

Energy of the nth state can expressed as

$\frac{13.6}{n^2}=1.507$

$\therefore n=\sqrt{\frac{13.6}{1.507}}=\sqrt{9.024}\approx 3$