Question

A group of hydrogen atoms are prepared in n = 4 states. List the wavelength that are emitted as the atoms make transitions and return to n = 2 states.

Answer 1

There will be three wavelengths.
(i) For the transition from (n = 4) to (n = 3) state
(ii) For the transition from (n = 3) to (n = 2) state
(iii) For the transition from (n = 4) to (n = 2) state

Let ($\lambda$₁) be the wavelength when the atom makes transition from (n = 4) state to (n = 2) state._​
Here,
n₁ = 2
n₂ = 4
Now, the wavelength ($\lambda$₁)​ will be
$\frac{1}{{\lambda }_1}=R\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$
$$\begin{gathered} R=1.097\times {10}^7{\mathrm{m}}^{-1} \\ \frac{1}{{\lambda }_1}=1.097\times {10}^7\times \left(\frac{1}{4}-\frac{1}{16}\right) \\ \Rightarrow \frac{1}{{\lambda }_1}=1.097\times {10}^7\left(\frac{4-1}{16}\right) \\ \Rightarrow \frac{1}{{\lambda }_1}=\frac{1.097\times {10}^7\times 3}{16} \\ \Rightarrow {\lambda }_1=\frac{16\times {10}^{-7}}{3\times 1.097} \\ =4.8617\times {10}^{-7} \\ =486.1\times {10}^{-9} \\ =487\mathrm{nm} \end{gathered}$$

When an atom makes transition from (n = 4) to (n = 3), the wavelength ($\lambda$2) is given by

$$\begin{gathered} \mathrm{Here}\mathrm{again}, \\ {n}_1=3 \\ {n}_2=4 \\ \frac{1}{{\lambda }_2}=1.097\times {10}^7\left(\frac{1}{9}-\frac{1}{16}\right) \\ \Rightarrow \frac{1}{{\lambda }_2}=1.097\times {10}^7\left(\frac{16-9}{144}\right) \\ \Rightarrow \frac{1}{{\lambda }_2}=\frac{1.097\times {10}^7\times 7}{144} \\ \Rightarrow {\lambda }_2=\frac{144}{1.097\times {10}^7\times 7} \\ =1875\mathrm{nm} \end{gathered}$$

Similarly, wavelength ($\lambda$₃) for the transition from (n = 3) to (n = 2) is given by
When the transition is n₁ = 2 to n₂ = 3:
$$\begin{gathered} \frac{1}{{\lambda }_3}=1.097\times {10}^7\left(\frac{1}{4}-\frac{1}{9}\right) \\ \Rightarrow \frac{1}{{\lambda }_3}=1.097\times {10}^7\left(\frac{9-4}{36}\right) \\ \Rightarrow \frac{1}{{\lambda }_3}=\frac{1.097\times {10}^7\times 5}{36} \\ \Rightarrow {\lambda }_3=\frac{36\times {10}^{-7}}{1.097\times 5}=656\mathrm{nm} \end{gathered}$$