Question

The radius of the second Bohr orbit for the hydrogen atom is : $$\begin{gathered} (\mathrm{Planck}’\mathrm{s}\mathrm{Constant},\mathrm{h}=6.6262\times {10}^{-34}\mathrm{Js};\mathrm{the}\mathrm{mass}\mathrm{of}\mathrm{electron}=9.1091\times {10}^{-31}\mathrm{kg};\mathrm{charge}\mathrm{of}\mathrm{electron},\mathrm{e}=1.60210\times {10}^{-19}\mathrm{C}; \\ \mathrm{permittivity}\mathrm{of}\mathrm{vacuum},\in 0=8.854185\times {10}^{-12}{\mathrm{kg}}^{-1}{\mathrm{m}}^{-3}{\mathrm{A}}^2) \end{gathered}$$

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

Answer 1

The correct option is B

$2.12\mathrm{angstorm}$

Explanation for the correct option:

B: $2.12\mathrm{angstorm}$

The expression used for the calculation of the radius of second Bohr's orbit for hydrogen

$\mathrm{r}=\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{4{\mathrm{\pi }}^2{\mathrm{me}}^2\mathrm{k}}$

Putting the values in the above expression

$\begin{array}{rcl}\mathrm{r}& =& \frac{2^2{(6.6262\times {10}^{-34}\mathrm{Js})}^2}{4\times {(3.1416)}^2\times 9.1091\times {10}^{-31}\mathrm{kg}\times {(1.60210\times {10}^{-19}\mathrm{C})}^2\times 9\times {10}^9}\\ & =& 2.12\times {10}^{-10}\mathrm{m}\\ & =& 2.12\mathrm{angstorm}\end{array}$

Final Answer: Therefore, option B is the correct option.