Question

The radius of the third Bohr orbit is $4.77\text{ }\overset{\text{o}}{\mathop{\text{A}}}\,.$ The radius of the second orbit will be:

• A. $\text{1}\text{.16 }\overset{\text{o}}{\mathop{\text{A}}}\,$
• B. $\text{3}\text{.18 }\overset{\text{o}}{\mathop{\text{A}}}\,$
• C. $\text{2}\text{.12 }\overset{\text{o}}{\mathop{\text{A}}}\,$
• D. $\text{1}\text{.59 }\overset{\text{o}}{\mathop{\text{A}}}\,$

Answer 1

Answer: (C)

$\text{2}\text{.12 }\overset{\text{o}}{\mathop{\text{A}}}\,$

Answer 2

Bohrs radius for ${{n}^{th}}$ orbit is given by ${{r}_{n}}=\frac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}m{{e}^{2}}k\cdot z}$ $i.e.,$ ${{r}_{n}}\propto {{n}^{2}}$ Hence, $\frac{{{r}_{3}}}{{{r}_{2}}}={{\left( \frac{3}{2} \right)}^{2}}=\frac{9}{4}$ $\therefore$ ${{r}_{2}}=\frac{4}{9}\times {{r}_{3}}=\frac{4}{9}\times 4.77=2.12\,\overset{\text{o}}{\mathop{\text{A}}}\,$