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Posted by Ritik Saini 3 years, 10 months ago (12700063)
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Preeti Dabral 3 years ago (2983787)
- Energy of the ground state (n = 1) = – (ionization energy) = –13.6 eV
The wavelength of the incident radiation, {tex}\lambda{/tex} = 975 {tex}\mathop {\text{A}}\limits^{\text{o}} {/tex}
{tex}\therefore{/tex} The energy of the incident photon = hc/{tex}\lambda{/tex}
= {tex}\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{975 \times 10^{-10} \times 1.6 \times 10^{-19}}{/tex} = 12.75 eV
Let electron is exerted to nth orbit,
{tex}\Rightarrow{/tex} 12.75 = 13.6 {tex}\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right){/tex}
{tex}\Rightarrow{/tex} n = 4
The quantum transitions to the less excited states gives six possible lines as follows:
n = 4 : (4 {tex}\rightarrow{/tex} 3), (4 {tex}\rightarrow{/tex} 2), (4 {tex}\rightarrow{/tex} 1)
n = 3 : (3 {tex}\rightarrow{/tex} 2), (3 {tex}\rightarrow{/tex} 1)
n = 2 : (2 {tex}\rightarrow{/tex} 1)
- The longest wavelength emitted is for the transitions (4 {tex}\rightarrow{/tex} 3) where energy difference is minimum.
Emin = (E4 - E3) = 13.6 {tex}\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right){/tex} = 0.661 eV
Thus {tex}\lambda_{\max }=\frac{h c}{\mathrm{E}_{\min }}{/tex}
= {tex}\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{0.661 \times 1.6 \times 10^{-19}} \mathrm{m}{/tex}
{tex}\approx{/tex} 18807 {tex}\mathop {\text{A}}\limits^{\text{o}} {/tex}
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