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Q. If the radius of the first orbit of hydrogen atom is $5.29 \times 10^{-11} \,m$, the radius of the second orbit will be

Atoms

Solution:

Radius of $n^{\text {th }}$ orbit $=n^{2} r_{0}$
Radius of $2^{\text {nd }}$ orbit $=4 r_{0}$
$\left[r_{0} \approx 5.3 \times 10^{-11} m \right]$
$\therefore $ Hence radius needed $\approx 21.16 \times 10^{-11} m$