A) \[n=3\] to \[n=1\]
B) \[n=10\] to \[n=1\]
C) \[n=9\] to \[n=1\]
D) \[n=2\] to \[n=1\]
Correct Answer: A
Solution :
Wave number of spectral line in emission spectrum of hydrogen, \[\overline{v}={{R}_{H}}\left( \frac{1}{{{n}^{2}}}-\frac{1}{n_{2}^{2}} \right)\] …. (i) Given, \[\overline{v}=\frac{8}{9}{{R}_{H}}\] On putting the value of v in Eq. (i), we get \[\frac{8}{9}{{R}_{H}}={{R}_{H}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] \[\frac{8}{9}=\frac{1}{{{(1)}^{2}}}-\frac{1}{n_{2}^{2}}\] \[\frac{8}{9}-1=-\frac{1}{n_{2}^{2}}\] \[\frac{1}{3}-\frac{1}{{{n}_{2}}}\] \[\therefore \] \[{{n}_{2}}=3\] Hence, electron jumps from \[{{n}_{2}}=3\] to \[{{n}_{1}}=1\]
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