JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 3)
The radius of electron's second stationary orbit in Bohr's atom is R. The radius of 3rd orbit will be
2.25R
$3 \mathrm{R}$
$\frac{\mathrm{R}}{3}$
$9 \mathrm{R}$
Explanation
$r \propto \frac{n^{2}}{Z}$
$$ \begin{aligned} & \frac{r_{2 { nd }}}{r_{3 \mathrm{rd}}}=\left(\frac{n_{2}}{n_{3}}\right)^{2} \\\\ & \Rightarrow \frac{R}{r_{3 r d}}=\left(\frac{2}{3}\right)^{2} \\\\ & \Rightarrow r_{3 \mathrm{rd}}=\frac{9}{4} R \\\\ & =2.25 R \end{aligned} $$
$$ \begin{aligned} & \frac{r_{2 { nd }}}{r_{3 \mathrm{rd}}}=\left(\frac{n_{2}}{n_{3}}\right)^{2} \\\\ & \Rightarrow \frac{R}{r_{3 r d}}=\left(\frac{2}{3}\right)^{2} \\\\ & \Rightarrow r_{3 \mathrm{rd}}=\frac{9}{4} R \\\\ & =2.25 R \end{aligned} $$
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