Question

The radius of the second Bohr orbit for the hydrogen atom is:
[Given: Planck's constant. $\mathrm{h}=6.6262\times {10}^{-34}\mathrm{Js}$; mass of electron $=9.1091\times {10}^{-31}\mathrm{kg}$; charge of electron, $\mathrm{e}=1.60210\times {10}^{-19}\mathrm{C}$; permittivity of vacuum, ${\mathrm{\epsilon }}_0=8.854185\times {10}^{-12}{\mathrm{kg}}^{-1}{\mathrm{m}}^{-3}{\mathrm{A}}^2]$

• A. $4.76{\mathrm{A}}^{°}$
• B. $0.529{\mathrm{A}}^{°}$
• C. $2.12{\mathrm{A}}^{°}$
• D. $1.65{\mathrm{A}}^{°}$

Answer 1

The correct option is C

$2.12{\mathrm{A}}^{°}$

Explanation for correct option:

(C) $2.12{\mathrm{A}}^{°}$

Step 1: Formula for Bohr radius

$$\begin{gathered} \mathrm{r}=\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{4{\mathrm{\pi }}^2{\mathrm{me}}^2\mathrm{K}} \\ \mathrm{Where}, \\ \mathrm{n}=\mathrm{bohr}\mathrm{orbit} \\ \mathrm{h}=\mathrm{planck}\text{'}\mathrm{s}\mathrm{constant} \\ \mathrm{m}=\mathrm{mass}\mathrm{of}\mathrm{electron} \\ \mathrm{e}=\mathrm{charge}\mathrm{of}\mathrm{electron} \end{gathered}$$

Step 2: Substituting values :

• Since, it is a second Bohr orbit,

$$\begin{gathered} \mathrm{n}=2 \\ \mathrm{r}=\frac{2^2\mathrm{x}6.626\mathrm{x}{10}^{-34}\mathrm{Js}{)}^2}{4\times {(3.1416)}^2\times 9.1091\times {10}^{-31}\mathrm{kg}\times {(1.60210\times 10{}^{-19}\mathrm{C})}^2\times 9\times {10}^9} \\ \mathrm{r}=2.12\mathrm{x}{10}^{-10}\mathrm{m} \\ \mathrm{r}=2.12{\mathrm{A}}^{°} \end{gathered}$$

Explanation for incorrect option:

The radius of the second Bohr orbit for the hydrogen atom is $2.12{\mathrm{A}}^{°}$. Thus options A, B, and D are incorrect.

Hence, the correct option is (C) i.e., $2.12{\mathrm{A}}^{°}$