Question

The time of revolution of an electron around a nucleus of charge Ze in n^th 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

The Correct Option is B

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

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

\(T =\frac{2\pi r}{v}\)

Radius of n^th 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, \(T \propto  \frac{n^{2}}{Z^{2}}\)