A) \[\frac{4}{5}\]
B) \[\frac{5}{4}\]
C) \[\frac{4}{3}\]
D) \[\frac{3}{4}\]
Correct Answer: D
Solution :
| Lyman series, \[{{n}_{1}}=1\] | |
| For third line of Lyman series, \[{{n}_{2}}=1\] | |
| For hydrogen, \[Z=1\] | |
| \[{}_{\overline{V}H}=\frac{1}{\lambda }={{R}_{H}}{{Z}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] | |
| \[={{R}_{H}}{{(1)}^{2}}\left( \frac{1}{1}-\frac{1}{{{\left( 4 \right)}^{2}}} \right)=\frac{15}{16}{{R}_{H}}\] | ? (i) |
| For lithium, Z=3 |
| For first line of Balmer series, \[{{n}_{1}}=2,{{n}_{2}}=3\] |
| \[{{\overline{v}}_{Li}}={{R}_{H}}{{(3)}^{2}}\left( \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(3)}^{2}}} \right)\]\[={{R}_{H}}\times 9\times \frac{5}{36}=\frac{5}{4}{{R}_{H}}\] |
| On dividing equation (i) by (ii), we get |
| \[\frac{{{\overline{v}}_{H}}}{{{\overline{v}}_{Li}}}=\frac{\left( \frac{15}{16} \right){{R}_{H}}}{\left( \frac{5}{4} \right){{R}_{H}}}=\frac{15}{16}\times \frac{4}{5}=\frac{3}{4}\] |
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