Question

If the average life time of an excited state of hydrogen is of the order of ${10}^{-8}s$, then the number of revolutions an electron will make when it is in $n=2$ state before coming to ground state will be [Take $a_0=0.53\mathring{A}$ and all standard data if required]

• A. ${10}^7$
• B. $8.25\times {10}^6$
• C. $2.37\times {10}^5$
• D. None of these

Answer 1

The correct option is B $8.25\times {10}^6$

Here, $Z=1$ and $n=2$
$\Rightarrow \nu =\frac{6.6\times {10}^{15}}{8}=0.825\times {10}^{15}$
Time which it stays in $n=2$ is ${10}^{-8}s$
No. of revolutions made is $=0.825\times {10}^{15}\times {10}^{-8}=8.25\times {10}^6$