Question

The average life of an excited state of hydrogen atom is of the order of ${10}^{8}s$. The number of revolutions made by an electron when it is in state n = 2 and before it suffers a transition to state n = 1, are

• A. $2.28\times {10}^6$
• B. $22.8\times {10}^6$
• C. $8.23\times {10}^6$
• D. $2.82\times {10}^6$

Answer 1

The correct option is C

$8.23\times {10}^6$

Velocity of electron, $v_n=2.19\times {10}^6\frac{Z}{n}m{s}^{-1}$ (Z=1 for H)
The distance travelled by the electron in second Bohr’s orbit in ${10}^{-8}s$
$=\frac{2.19\times {10}^6\times 1.0\times {10}^{-8}}{2}m=1.095\times {10}^{-2}m$
The circumference of second orbit
$=2\pi r=2\pi \times (0.529\times \frac{n^2}{Z})=2\pi (0.529\times \frac{4}{1})=13.30\times {10}^{-10}m$
Number of revolutions =$\frac{1.095\times {10}^{-2}}{13.6\times {10}^{-10}}=8.23\times {10}^6$