A) n = 3 t on =1
B) n =10 to n = 1
C) n = 9 t on = 1
D) n=2 t on = 1
Correct Answer: A
Solution :
Wave number of spectral line in emission spectrum of hydrogen, \[\bar{v}={{R}_{H}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] Given, \[\bar{v}=\frac{8}{9}{{R}_{H}}\] On putting the value of v\[\bar{v}\]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\[{{\text{n}}_{\text{2}}}\text{=3to }{{\text{n}}_{\text{1}}}\text{=1}\]
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