Question

Calculate the number of revolutions/sec around the nucleus of an electron placed in III orbit of $H$-atom.

• A. $1.44\times {10}^{12}$
• B. $7.29\times {10}^7$
• C. $15.33\times {10}^{14}$
• D. $2.44\times {10}^{14}$

Answer 1

The correct option is D $2.44\times {10}^{14}$

As we know,
For C.G.S system $u_n=\sqrt{\left(\frac{Z{e}^2}{m{r}_n}\right)}$
For electron $\because e=4.803\times {10}^{-10}esu$
$m=9.108\times {10}^{-28}g$
Radius of III orbit$=r_1\times n^2=0.529\times {10}^{-8}\times 9cm$
$\therefore u_n=\sqrt{\left(\frac{1\times (4.803\times {10}^{-10}{)}^2}{9.108\times {10}^{-28}\times 0.529\times {10}^{-8}\times 9}\right)}$
$u_n=7.29\times {10}^{7}cm$ $se{c}^{-1}$
Now, circumference of III orbit$=2\times \pi \times 0.529\times {10}^{-8}\times 9$
$=29.93\times {10}^{-8}cm$
$\therefore$ No.of revolutions/sec$=\frac{u_n}{2\pi r}=\frac{7.29\times {10}^7}{29.93\times {10}^{-8}}$
$=2.44\times {10}^{14}$