A) \[\frac{27}{20}\lambda \]
B) \[\frac{20}{27}\lambda \]
C) \[\frac{27}{20}\lambda \]
D) none of these
Correct Answer: B
Solution :
The wavelength of series for n is given by \[\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{n}^{2}}} \right)\]where R is Rydberg's constant. For Balmer series\[n=3\]gives the first member of series and n = 4 gives the second member of series. Hence, \[\frac{1}{{{\lambda }_{1}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)\] \[\therefore \] \[\frac{1}{{{\lambda }_{1}}}=R\left( \frac{5}{36} \right)\] ? (i) \[\frac{1}{{{\lambda }_{2}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)\] \[=R\left( \frac{12}{16\times 4} \right)=\frac{3R}{16}\] ?. (ii) \[\Rightarrow \] \[\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{16}{3}\times \frac{5}{36}=\frac{20}{27}\] Given, \[{{\lambda }_{1}}=\lambda \]\[\Rightarrow \]\[{{\lambda }_{2}}=\frac{20}{27}\lambda \]
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