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Let $R_1$ be the radius of the second stationary orbit and $R_2$ be the radius of the fourth stationary orbit of an electron in Bohr's model. The ratio $\frac{R_1}{R_2}$ is:
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Verified Answer
The correct answer is:
0.25
As,
$\begin{aligned}
\mathrm{R}_n & =0.529 \frac{n^2}{\mathrm{Z}} \\
\text { Hence, } \quad \frac{\mathrm{R}_1}{\mathrm{R}_2} & =\frac{0.529 \frac{n_1^2}{\mathrm{Z}_1}}{0.529 \frac{n_2^2}{\mathrm{Z}_2}} \\
\frac{\mathrm{R}_1}{\mathrm{R}_2} & =\frac{n_1^2}{n_2^2}=\frac{(2)^2}{(4)^2} \\
\therefore \quad \frac{\mathrm{R}_1}{\mathrm{R}_2} & =\frac{1}{4}=0.25
\end{aligned}$
(where, the symbols have their usual meanings)
$\begin{aligned}
\mathrm{R}_n & =0.529 \frac{n^2}{\mathrm{Z}} \\
\text { Hence, } \quad \frac{\mathrm{R}_1}{\mathrm{R}_2} & =\frac{0.529 \frac{n_1^2}{\mathrm{Z}_1}}{0.529 \frac{n_2^2}{\mathrm{Z}_2}} \\
\frac{\mathrm{R}_1}{\mathrm{R}_2} & =\frac{n_1^2}{n_2^2}=\frac{(2)^2}{(4)^2} \\
\therefore \quad \frac{\mathrm{R}_1}{\mathrm{R}_2} & =\frac{1}{4}=0.25
\end{aligned}$
(where, the symbols have their usual meanings)
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