Question

The average life of an excited state of hydrogen atom is of the order of ${10}^{-8}sec$. The number of revolutions made by an electron when it returns from $n=2$ to $n=1$ is:

• 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 $8.23\times {10}^6$

$V_n=2.18\times {10}^6\frac{z}{n}㎧\therefore V$ of $2^{nd}$ orbit $=\frac{2.18\times {10}^6}{2}$

$\therefore$ distance traveled in ${10}^{-8}s=\frac{2.18\times {10}^6}{2}\times {10}^{-8}=1.09\times {10}^{-2}$

Number of revolution$=\frac{distance}{circumference}=\frac{1.09\times {10}^{-2}}{0.53\times {10}^{-10}\times 4\times 2\pi }$

$\simeq 8.23\times {10}^6$ revolution

i.e., option C