A) 4:1
B) 1: 4
C) 27 : 5
D) 5: 27
Correct Answer: B
Solution :
Shortest wavelength in a spectral series corresponds to series limit i.e., \[{{n}_{2}}=\infty \] For series limit of Balmer series \[{{n}_{1}}=2\] and \[{{n}_{2}}=\infty \] So, \[\frac{1}{{{\lambda }_{1}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)=\frac{R}{4}\] Since, \[c=\frac{c}{\lambda }\] so, \[\frac{1}{\lambda }=\frac{v}{c}\] so, \[\frac{{{v}_{1}}}{c}=\frac{R}{4}\]?(i) Similarly, for series limit of Lyman series \[{{n}_{1}}=1\] and \[{{n}_{2}}=\infty \] So, \[\frac{1}{{{\lambda }_{2}}}=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)=\frac{R}{1}\] or \[\frac{{{v}_{2}}}{c}=\frac{R}{1}\] ?.(ii) From equation (i) and (ii) we get \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{R/4}{R/1}=\frac{1}{4}\] So, \[{{v}_{1}}:{{v}_{2}}=1:4\]
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