Hydrogen atom in its ground state is excited by a radiation of wavelength 97.26 nm. The longest wavelength it emits is

Hydrogen atom in its ground state is excited by a radiation of wavelength 97.26 nm. The longest wavelength it emits is

1. A

1250 nm

2. B

1400 nm

3. C

1875 nm

4. D

2050 nm

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Solution:

$\mathrm{\Delta }E=\frac{hc}{\lambda }=\frac{\left(6.626×{10}^{-34}\mathrm{Js}\right)\left(3×{10}^{8}{\mathrm{ms}}^{-1}\right)}{\left(97.26×{10}^{-9}\mathrm{m}\right)}=2.0438×{10}^{-18}\mathrm{J}$

Also $\mathrm{\Delta }E=\left(2.18×{10}^{-18}\mathrm{J}\right)\left(\frac{1}{{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)$

Hence $\begin{array}{l}\frac{1}{{n}_{2}^{2}}=1-\left(\frac{\mathrm{\Delta }E}{2.18×{10}^{-18}\mathrm{J}}\right)=1-\frac{2.0436×{10}^{-18}}{2.18×{10}^{-18}}=1-0.9374\\ {n}_{2}=\sqrt{1/\left(1-0.9374\right)}\simeq 4\end{array}$

The transition ${n}_{2}=4\to {n}_{1}=3$ will emit the longest wavelength. Hence

$\begin{array}{l}\mathrm{\Delta }E=\left(2.18×{10}^{-18}\mathrm{J}\right)\left(\frac{1}{{3}^{2}}-\frac{1}{{4}^{2}}\right)=1.06×{10}^{-19}\mathrm{J}\\ \lambda =\frac{hc}{\mathrm{\Delta }E}=\frac{\left(6.626×{10}^{-34}\mathrm{Js}\right)\left(3×{10}^{8}{\mathrm{ms}}^{-1}\right)}{\left(1.06×{10}^{-19}\mathrm{J}\right)}=1.875×{10}^{-6}\mathrm{m}=1875\mathrm{nm}\end{array}$