Question

Find out the longest wavelength of absorption line for hydrogen gas-containing atoms in the ground state.

Answer 1

Calculating Wavelength$\left(\mathrm{\lambda }\right)$:

1. As the transition of electron takes place from the ground state, so ${\mathrm{n}}_1=1$
2. From the formula: $\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$; here $\mathrm{Z}$(atomic number) for hydrogen = $1$
3. we will take $n_2=2and\infty$ to check the largest wavelength.
4. $$\begin{gathered} Taking{n}_2=\infty \\ \Rightarrow \frac{1}{\lambda }=R{Z}^2\left(\frac{1}{1^2}-\frac{1}{\infty }\right) \\ \Rightarrow \frac{1}{\lambda }=R{\left(1\right)}^2\times 1 \\ \Rightarrow \lambda =\frac{1}{R} \end{gathered}$$
5. ${n}_{2}=2\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\lambda }=R{\left(1\right)}^{2}(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}})\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\lambda }=R(\frac{1}{1}-\frac{1}{4})\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\lambda }=R\frac{3}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \lambda =\frac{4R}{3}$$$\begin{gathered} Taking{n}_2=2 \\ \Rightarrow \frac{1}{\lambda }=R{Z}^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \\ \Rightarrow \frac{1}{\lambda }=R{\left(1\right)}^2\left(\frac{1}{1}-\frac{1}{4}\right)\Rightarrow \frac{1}{\lambda }=R\frac{3}{4} \\ \Rightarrow \lambda =\frac{4}{3R} \end{gathered}$$​
6. We can see that $\mathrm{\lambda }$ is maximum when $\mathrm{n}=2$.
7. So maximum value of wavelength will be $\lambda =\frac{4}{3\times 109673}=1.215\times {10}^{-5}cm=121.5nm$
8. We can conclude that in order to have a high wavelength the change in energy should be minimum. i.e. we got maximum wavelength when electron transitioned from $\mathrm{n}=1\mathrm{to}\mathrm{n}=2$.