The approximate value of principal quantum number for a circular orbit of hydrogen atom of radius $530…
The approximate value of principal quantum number for a circular orbit of hydrogen atom of radius $530 \mathrm{~nm}$ is
- 26
- 100
- 200
- 21
Solution
Radius of electron of hydrogen in $n^{\text {th }}$ orbit,
$\begin{aligned}
& r_n=n^2 \times 0.53 Å \\
& 530 \mathrm{~nm}=n^2 \times 0.053 Å \\
& \therefore \quad 530 \times 10^{-9} \mathrm{~m}=n^2 \times 0.53 \times 10^{-10} \mathrm{~m} \\
& \Rightarrow \quad n^2=\frac{5300}{0.53} \\
& \Rightarrow \quad n^2=10^4 \\
& n=100 \\
&
\end{aligned}$
Asked in: AP EAMCET 2018 (24 Apr Shift 1)
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