A) \[n=4\]to\[n=3\]
B) \[n=3\]to\[n=2\]
C) \[n=4\] to\[n=2\]
D) \[n=3\]to \[n=1\]
E) \[~n=2\] to\[n=1\]
Correct Answer: E
Solution :
According to Rydbergs formula \[\frac{1}{\lambda }=R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right){{Z}^{2}}\] For\[H{{e}^{+}}\]ion;\[Z=2,{{n}_{1}}=2\]and \[{{n}_{2}}=4\] \[\therefore \] \[\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)\times {{2}^{2}}\] \[=R\times \frac{3}{16}\times 4\] \[\frac{1}{\lambda }=\frac{3R}{4}\] Or \[\lambda =\frac{4}{3R}\] For getting the same value of wavelength in hydrogen atomic spectrum, the values of\[{{n}_{1}}\]and\[{{n}_{2}}\]must be 1 and 2 respectively. \[\therefore \] \[\frac{1}{\lambda }=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)\times {{1}^{2}}\] \[\frac{1}{\lambda }=\frac{3R}{4}\] Or \[\lambda =\frac{4}{3R}\]
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