Question

The velocity of an electron in the first Bohr orbit of a hydrogen atom is $2.19\times {10}^{6}m{s}^{-1}$. Its velocity in the second orbit would be:

• A. $1.10\times {10}^{6}m{s}^{-1}$
• B. $4.38\times {10}^{6}m{s}^{-1}$
• C. $5.5\times {10}^{5}m{s}^{-1}$
• D. $8.76\times {10}^{6}m{s}^{-1}$

Answer 1

The correct option is A $1.10\times {10}^{6}m{s}^{-1}$

$v_n=\frac{2.19\times {10}^6\times Z}{n}m/sec$.
For $n=2$ of hydrogen atom,
$v_2=\frac{2.19\times {10}^6\times m/sec}{2}=1.09\times {10}^{6}m/sec.$