A) \[\frac{1}{2}\]
B) \[\frac{2}{1}\]
C) \[\frac{5}{4}\]
D) \[\frac{3}{4}\]
Correct Answer: D
Solution :
Number of spectral lines obtained due to transition of electrons from nth orbit to lower orbit is, \[N=\frac{n(n-1)}{2}\] I case \[6=\frac{{{n}_{1}}({{n}_{1}}-1)}{2}\]\[\Rightarrow \]\[{{n}_{1}}=4\] II case \[3=\frac{{{n}_{2}}({{n}_{2}}-1)}{2}\]\[\Rightarrow \]\[{{n}_{2}}=3\] Velocity of electron in hydrogen atom in nth orbit \[{{v}_{n}}\propto \frac{1}{n}\] \[\frac{{{v}_{n}}}{v{{}_{n}}}=\frac{{{n}_{2}}}{{{n}_{1}}}\] \[\Rightarrow \] \[\frac{{{v}_{6}}}{{{v}_{3}}}=\frac{3}{4}\]
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