# Question #81091

##### 1 Answer
Dec 30, 2017

From Bohr's model of atom we know that the electron is held in a circular orbit by electrostatic attraction. In case of equilibrium the centripetal force is equal to the Coulomb force.

$\frac{{m}_{e} {v}^{2}}{r} = \frac{Z {k}_{e} {e}^{2}}{r} ^ 2$
where ${m}_{e}$ is mass of electron, $e$ is the charge of electron, ${k}_{e}$ is Coulomb's constant and $Z$ is the atomic number of atom and $r$ is radius of orbit.

For Hydrogen $Z = 1$, therefore above equation reduces to

$\frac{{m}_{e} {v}^{2}}{r} = \frac{{k}_{e} {e}^{2}}{r} ^ 2$

Rewriting above in terms of Kinetic energy of electron we have

$K E = \frac{1}{2} {m}_{e} {v}^{2} = \frac{1}{2} \frac{{k}_{e} {e}^{2}}{r}$ ......(1)

From Coulomb's Law have have potential energy of the system as

$P E = - \frac{{k}_{e} {e}^{2}}{r}$ .......(2)

Comparing (1) and (2) we see that

$P E = - 2 K E$ .....(3)

and Total energy

$T E = P E + K E = - K E$ ..........(4)

To check which of the given (4) statements are true you need to remember (3) and (4) and test each statement.

Also see that potential energy in ground state is taken to be zero. Potential energy of a system is dependent on the reference. With the change of reference, both $P E \mathmr{and} T E$ change.