The radius of the orbit of an electron in a H | vedclass

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(C) The radius of the orbit is given by $r_n = a_0 \frac{n^2}{Z} = 4.5 a_0$. Given orbital angular momentum $L = \frac{nh}{2\pi} = \frac{3h}{2\pi}$,so

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$(A, C)$

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(C) The radius of the orbit is given by $r_n = a_0 \frac{n^2}{Z} = 4.5 a_0$. Given orbital angular momentum $L = \frac{nh}{2\pi} = \frac{3h}{2\pi}$,so $n = 3$. Substituting $n=3$ into the radius formula: $4.5 = \frac{3^2}{Z} = \frac{9}{Z}$,which gives $Z = 2$. The atom is a Helium ion $(He^+)$. When the atom de-excites from $n=3$,the possible transitions are $3 \rightarrow 2$,$3 \rightarrow 1$,and $2 \rightarrow 1$. The wavelength $\lambda$ is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$. For $3 \rightarrow 1$: $\frac{1}{\lambda_{3 \rightarrow 1}} = R(2^2) \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 4R \left( 1 - \frac{1}{9} \right) = 4R \left( \frac{8}{9} \right) = \frac{32R}{9} \Rightarrow \lambda_{3 \rightarrow 1} = \frac{9}{32R}$. For $3 \rightarrow 2$: $\frac{1}{\lambda_{3 \rightarrow 2}} = R(2^2) \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{9} \right) = 4R \left( \frac{5}{36} \right) = \frac{5R}{9} \Rightarrow \lambda_{3 \rightarrow 2} = \frac{9}{5R}$. For $2 \rightarrow 1$: $\frac{1}{\lambda_{2 \rightarrow 1}} = R(2^2) \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 4R \left( 1 - \frac{1}{4} \right) = 4R \left( \frac{3}{4} \right) = 3R \Rightarrow \lambda_{2 \rightarrow 1} = \frac{1}{3R}$. The possible wavelengths are $\frac{9}{32R}$ and $\frac{9}{5R}$. Thus,the correct option is $(C)$.

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