A) \[\frac{4}{9}\]
B) \[\frac{9}{4}\]
C) \[\frac{27}{5}\]
D) \[\frac{5}{27}\]
Correct Answer: D
Solution :
| In hydrogen atom. Wavelength of characteristic spectrum |
| \[\frac{1}{\lambda }=R{{z}^{2}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] |
| For Lyman series \[{{n}_{1}}=1,{{n}_{2}}=2\] |
| \[\frac{1}{{{\lambda }_{1}}}=R{{z}^{2}}\left[ \frac{1}{{{(1)}^{2}}}-\frac{1}{{{(2)}^{2}}} \right]\,\] (i) |
| For Balmer series \[{{n}_{1}}=2,{{n}_{2}}=3\] |
| \[\frac{1}{{{\lambda }_{2}}}=R{{z}^{2}}\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(3)}^{2}}} \right]\,\] (ii) |
| Dividing Eq. (ii) by Eq. (i) we get |
| \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{R{{z}^{2}}\left[ \frac{1}{4}-\frac{1}{9} \right]}{R{{z}^{2}}\left[ 1-\frac{1}{4} \right]}=\frac{\frac{5}{36}}{\frac{3}{4}}\] |
| \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{5}{36}\times \frac{4}{3}=\frac{5}{27}\] |
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