Question

The time of revolution of an electron around a nucleus of charge Ze in nth Bohr orbit is directly proportional to

• A. $n$
• B. $\frac{n^{3}}{Z^{2}}$
• C. $\frac{n^{2}}{Z}$
• D. $\frac{Z}{n}$

Answer 1

Answer: (B)

$\frac{n^{3}}{Z^{2}}$

Answer 2

The correct option is (B): $\frac{n^{3}}{Z^{2}}$

The time of revolution of an electron around a nucleus of charge Ze in nth Bohr orbit is directly proportional to

$T =\frac{2\pi r}{v}$

Radius of nth orbit = $\frac{n^{2}h^{2}}{\pi mZe^{2}}$

v = speed of electron in nth orbit = $\frac{Ze^{2}}{2 \epsilon _{0}nh}$

So, $T = \frac{4 \epsilon_{0}^{2}n^{3}h^{3}}{mZ^{2}e^{4}}$

So, $T \propto \frac{n^{2}}{Z^{2}}$