A) \[912\overset{o}{\mathop{A}}\,\]
B) \[304\overset{o}{\mathop{A}}\,\]
C) \[229\overset{o}{\mathop{A}}\,\]
D) \[687\overset{o}{\mathop{A}}\,\]
Correct Answer: B
Solution :
As, \[\Delta {{\varepsilon }_{12}}=\frac{hc}{{{h}_{\max }}}\] For \[He,\] \[z=2,\] and for longest wavelength \[{{n}_{1}}=1\] and \[{{n}_{2}}=2\] Now, \[\frac{1}{\lambda }=Z_{R}^{2}\,\,\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] \[=4\times 1.097\times {{10}^{7}}\left( 1-\frac{1}{4} \right)\] \[=4\times 1.097\times {{10}^{7}}\times \frac{3}{4}\] \[\frac{1}{\lambda }=3.29\times {{10}^{7}}/m\] So, \[\lambda =\frac{1}{3.29\times {{10}^{7}}}\] \[\lambda =0.3039\times {{10}^{-7}}m=304\overset{o}{\mathop{A}}\,\]
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