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 the radius of the first orbit of a hydrogen atom $\left(a_0\right)$$as0.53\mathrm{\AA \; and\; all\; standard\; data\; if\; required]}$

• A. ${10}^9$
• B. $8\times {10}^6$
• C. 2 $\times {10}^3$
• D. 5× ${10}^{10}$

Answer 1

The correct option is B $8\times {10}^6$

Time period for nth energy level electron is,
$T=\frac{2\pi r_n}{v_n}=\frac{4{\pi }^{2}m}{h}\times \frac{{r_n}^2}{n}$
$r_n=n^2{a}_0$
$T=\frac{4{\pi }^{2}m}{h}\times n^3{a_0}^2$
Required number of revolutions, $N=\frac{{10}^{-8}}{T}$
After substituting $n=2,m=9.1\times {10}^{-31}\mathrm{kg},$
and $h=6.63\times {10}^{-34}J-s,$$\mathrm{we}\mathrm{get}N=8\times {10}^6$