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 $K = \frac{1}{2} m v^{2}$ of the electron in C.G.S. system is equal to?

\frac{1}{2} \frac{Z e^{2}}{r}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{2} \frac{Z e^{2}}{r^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{Z e^{2}}{r}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{Z e^{2}}{r^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{1}{2} \frac{Z e^{2}}{r}$
In the revolution of an electron, coulomb force provides the necessary centripetal force

$\Rightarrow \frac{z e^{2}}{r^{2}} = \frac{m v^{2}}{r} \Rightarrow m v^{2} = \frac{z e^{2}}{r}$

$∴ K . E . \frac{1}{2} m v^{2} = \frac{Z e^{2}}{2 r}$

Suggest Corrections

Similar questions

Q. According to Bohr's postulate, electrons revolve around the nucleus in certain definite circular paths called stationary orbits in which they do not radiate energy.

Q. Select the correct curve(s).
If v = velocity of electron in Bohr's orbit
r = Radius of electron n Bohr's orbit
P.E = Potential energy of electron in Bohr's orbit
K.E = Kinetic energy of electron in Bohr's orbit

Q. A satellite moves around the earth in a circular orbit of radius

r

with speed

v

. If the mass of the satellite is

M

, its total energy is

In hydrogen atom electron of charge $- e$ and mass m revolves round the nucleus in a circular orbit of radius r. The electrostatic potential energy of the electron is $\frac{1}{4 π ε_{0}}$ times

Q. An electronic charge

e

is revolving in a circular orbit of radius

r

around a nucleus with speed

v

. The equivalent current is

Join BYJU'S Learning Program

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=12mv2 of the electron in C.G.S. system is equal to?

The correct option is A 12Ze2rIn the revolution of an electron, coulomb force provides the necessary centripetal force⇒ze2r2=mv2r⇒mv2=ze2r∴K.E.12mv2=Ze22r

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 $K = \frac{1}{2} m v^{2}$ of the electron in C.G.S. system is equal to?

The correct option is A $\frac{1}{2} \frac{Z e^{2}}{r}$
In the revolution of an electron, coulomb force provides the necessary centripetal force

$\Rightarrow \frac{z e^{2}}{r^{2}} = \frac{m v^{2}}{r} \Rightarrow m v^{2} = \frac{z e^{2}}{r}$

$∴ K . E . \frac{1}{2} m v^{2} = \frac{Z e^{2}}{2 r}$