A) \[\frac{1}{2}\]
B) \[\frac{2}{1}\]
C) \[\frac{5}{4}\]
D) \[\frac{3}{4}\]
Correct Answer: D
Solution :
Number of spectrical lines obtained due to transition of electrons from n^ orbit to lower orbit is given by \[N=\frac{n(n-1)}{2}\] Case-I: \[6=\frac{{{n}_{1}}({{n}_{1}}-1)}{2}\,\Rightarrow \,{{n}_{1}}=3\] Case-II: \[3=\frac{{{n}_{2}}({{n}_{2}}-1)}{2}\Rightarrow {{n}_{2}}=3\] Velocity of electron in hydrogen atom in nth orbit \[{{v}_{n}}\propto \,\frac{1}{n}\Rightarrow \,\frac{{{v}_{n}}}{{{v}_{n}}}=\frac{{{n}_{2}}}{{{n}_{1}}}\,\Rightarrow \,\frac{{{v}_{6}}}{{{v}_{3}}}\,=\frac{3}{4}\]You need to login to perform this action.
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