A) 2
B) 5
C) 3
D) 6
Correct Answer: A
Solution :
[a] For the third line of Balmer series, \[{{n}_{1}}=2,{{n}_{2}}=5\] \[\therefore \frac{1}{\lambda }=R{{Z}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)=R{{Z}^{2}}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{5}^{2}}} \right)=\frac{21R{{Z}^{2}}}{100}\] \[E=-13.6eV\] \[{{Z}^{2}}\times \frac{21}{100}=\frac{hc}{\lambda }=\frac{1242eVnm}{108.5nm}\] \[{{Z}^{2}}=\frac{1242\times 100}{108.5\times 21\times 13.6}=4\Rightarrow Z=2\]
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