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The radius of the first Bohr orbit of a hydrogen atom is $0.53 \times 10^{-8} \mathrm{~cm}$. The velocity of the electron in the first Bohr orbit is
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Verified Answer
The correct answer is:
$2.188 \times 10^{9} \mathrm{~cm} \mathrm{~s}^{-1}$
Hint:
$r=0.53 \times 10^{-9} \mathrm{~cm}=0.53 \times 10^{-9} n^{2}$
$\therefore \mathrm{n}=1$
$\mathrm{~V}=2.18 \times 10^{9} \mathrm{cm} / \mathrm{s}$
$r=0.53 \times 10^{-9} \mathrm{~cm}=0.53 \times 10^{-9} n^{2}$
$\therefore \mathrm{n}=1$
$\mathrm{~V}=2.18 \times 10^{9} \mathrm{cm} / \mathrm{s}$
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