Question

According to the Bohr theory for the hydrgen atom, the number of revolutions of the electron per second in the orbit of quantum number, n is proportional to :

• A. $n^{-2}$
• B. $\sqrt{n}$
• C. $n^{-3}$
• D. $n^{-1}$

Answer 1

Answer: (C)

$n^{-3}$

Number of revolution per sec $=\frac{v}{2\pi r}$
$=\frac{2.18\times {10}^6\times Z/n}{2\times 3.14\times (0.529\times \frac{n^2}{Z})}$
$\propto \frac{Z^2}{n^3}$

Number of revolution per $\propto n^{-3}$