A) 7/5
B) 27/20
C) 27/5
D) 20/7
Correct Answer: D
Solution :
| Here, for wavelength \[{{\lambda }_{1}}\] |
| \[{{n}_{1}}=4\] and \[{{n}_{2}}=3\] |
| and for \[{{\lambda }_{2}},\,{{n}_{1}}=3\] and \[{{n}_{2}}=2\] |
| We have \[\frac{hc}{\lambda }=-13.6\left[ \frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}} \right]\] |
| So, for \[{{\lambda }_{1}}\] |
| \[\Rightarrow \] \[\frac{hc}{{{\lambda }_{1}}}=-13.6\left[ \frac{1}{{{(4)}^{2}}}-\frac{1}{{{(3)}^{2}}} \right]\] |
| \[\frac{hc}{{{\lambda }_{1}}}=13.6\left[ \frac{7}{144} \right]\] (i) |
| Similarly, for \[{{\lambda }_{2}}\] |
| \[\Rightarrow \] \[\frac{hc}{{{\lambda }_{2}}}=-13.6\left[ \frac{1}{{{(3)}^{2}}}-\frac{1}{{{(2)}^{2}}} \right]\] |
| \[\frac{hc}{{{\lambda }_{2}}}=13.6\left[ \frac{5}{36} \right]\] (ii) |
| Hence, from Eqs. (i) and (ii), we get |
| \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{20}{7}\] |
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