Question

If m is mass of electron, v its velocity, r the radius of stationary circular orbit around a nucleus with charge Ze, then from Bohr's first postulate, the kinetic energy $K = \frac{1}{2}mv^{2}$of the electron in C.G.S. system is equal to.

• A. $\frac{1}{2}\frac{Ze^{2}}{r}$
• B. $\frac{1}{2}\frac{Ze^{2}}{r^{2}}$
• C. $\frac{Ze^{2}}{r}$
• D. $\frac{Ze}{r^{2}}$

Answer 1

Answer: (A)

$\frac{1}{2}\frac{Ze^{2}}{r}$

Answer 2

In the revolution of electron, coulomb force provides the necessary centripetal force

⇒ $\frac{ze^{2}}{r^{2}} = \frac{mv^{2}}{r}$ ⇒ $mv^{2} = \frac{ze^{2}}{r}$

∴ K.E.$= \frac{1}{2}mv^{2} = \frac{ze^{2}}{2r}$.