Question

If the average life time of an excited state of $H$ atom is of order ${10}^{-8}$sec, then the no. of orbits an $e^{-}$ makes when it is in the state $n=2$ and before it suffers a transition to $n=1$ are:

• A. $8\times {10}^6$
• B. $9\times {10}^6$
• C. $7\times {10}^6$
• D. $6\times {10}^6$

Answer 1

Answer: (A)

$8\times {10}^6$

The velocity of the electron in the second orbit of H atom is $u=1.09\times {10}^{8}cm/s$.
The orbital frequency is $\frac{velocityofelectron}{circumferenceoforbit}=\frac{u}{2\pi r}=\frac{1.09\times {10}^{8}cmsec/s}{2\times 3.1416\times 4\times 0.529\times {10}^{-8}cm}=8.2\times {10}^{14}/s$
The number of orbits made by the electron $=8.2\times {10}^{14}/s\times {10}^{-8}s=8.19\times {10}^6\approx 8\times {10}^6$.

Hence, the correct option is $A$