Question

The radius of the second Bohr orbit for the hydrogen atom is:

[Given: Planck's const. $h=6.6262\times {10}^{-34}Js$; mass of electron $=9.1091\times {10}^{-31}kg$; charge of electron, $e=1.60210\times {10}^{-19}C$; permittivity of vacuum, ${\epsilon }_0=8.854185\times {10}^{-12}k{g}^{-1}{m}^{-3}{A}^2$]

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

Answer 1

The correct option is C $2.12\mathring{A}$

$r=\frac{n^2{h}^2}{4{\pi }^{2}m{e}^{2}k}$

$r=\frac{2^2(6.6262\times {10}^{-34}Js{)}^2}{4\times (3.1416{)}^2\times 9.1091\times {10}^{-31}kg\times (1.60210\times {10}^{-19}C{)}^2\times 9\times {10}^9}$

$r=2.12\times {10}^{-10}m$

$r=2.12\mathring{A}$

Hence, the option (C) is the correct answer.