A) \[2R\]
B) \[3R\]
C) \[27R\]
D) \[9R\]
Correct Answer: D
Solution :
: As per question, \[\frac{{{\upsilon }_{h}}}{{{\upsilon }_{g}}}=\frac{1}{3}\] ??..(i) where subscripts h and g denotes higher energy state and ground state. Orbital velocity of electron^ in the nth orbit is \[{{\upsilon }_{n}}=\frac{{{e}^{2}}}{2{{\varepsilon }_{0}}nh}\] or \[{{\upsilon }_{n}}\propto \frac{1}{n}\] For ground state, \[n=1\] \[\therefore \] \[\frac{{{\upsilon }_{h}}}{{{\upsilon }_{g}}}=\frac{1}{n}\] ?...(ii) Equating (i) and (ii), we get \[n=3\] Radius of nth orbit is \[{{r}_{n}}=\frac{{{n}^{2}}{{h}^{2}}{{\varepsilon }_{0}}}{\pi {{e}^{2}}m}\] or \[{{r}_{n}}\propto {{n}^{2}}\] \[\therefore \] \[\frac{{{r}_{3}}}{{{r}_{1}}}=\frac{{{(3)}^{2}}}{{{(1)}^{2}}}=9\] \[{{r}_{3}}=9{{r}_{1}}=9R\](\[\because \] \[{{r}_{1}}=R\](Given))
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