The transition from the n = 7 to the n = 2 level of the hydrogen atom is accompanied by the emission of radiation slightly beyond the range of human perception, in the ultraviolet region. Determine the energy and wavelength.
Since spectral line in the ultraviolet region of spectrum and \(n_{h}=7\)
$$ \begin{aligned} &E=R_{H}\left(\frac{1}{2^{2}}-\frac{1}{n_{h}^{2}}\right) \\ &\text { where } R_{H}=1.097 \times 10^{7} \mathrm{~m}^{-1} \\ &=2.179 \times 10^{-18} J \end{aligned} $$
when \(n_{h}=7\).
$$ \begin{aligned} &E=2.179 \times 10^{-18} J\left(\frac{1}{2^{2}}-\frac{1}{7^{2}}\right) \\ &E=2.179 \times 10^{-18} J \times \frac{45}{196} \\ &E=5.003 \times 10^{-19} J \end{aligned} $$
\(\because \lambda=\frac{h c}{E}\)
\(\lambda=\frac{6.626 \times 10^{-34} J_{S} \times 2.998 \times 10^{8} \mathrm{~ms}^{-1}}{5.003 \times 10^{-19} J}\)
\(\lambda=3.971 \times 10^{-7} \mathrm{~m}\)
\(=3.971 \times 10^{-7} \mathscr{h}\left(\frac{1 \mathrm{~nm}}{10^{-9}{\mathrm{~m}}}\right)\)
\(=397.1 \mathrm{~nm}\)