Question

The approximate value of quantum number $n$ for the circular orbit of hydrogen $0.0001mm$ diameter is

• A. $1000$
• B. $60$
• C. $10000$
• D. $31$

Answer 1

The correct option is D $31$

$r=0.528n^2\times {10}^{-10}$

$\left(\frac{1\times {10}^{-7}}{2}\right)=0.528n^2\times {10}^{-10}$

$\Rightarrow n^2=\frac{5\times {10}^{-8}}{0.528\times {10}^{-10}}\Rightarrow n\approx 31$