Question

On the basis of Bohr's postulates derive an expression for orbital velocity of an electron in $n$th stationary orbit of hydrogen atom.
The ground state energy of hydrogen atom is $(-X)eV$. What will be the kinetic energy of the electron in this state?

Answer 1

From Bohr's postulates, angular momentum of an electron in $n^{th}$ orbit is given by:

$L_n=\frac{nh}{2\pi }$

$\therefore m_e{v}_n{r}_n=\frac{nh}{2\pi }................\left(i\right)$

From Coulomb's Law,

$F=\frac{Q_1{Q}_2}{4\pi {\epsilon }_o{r}^2}$

$F=\frac{e^2}{4\pi {\epsilon }_o{r}_n^2}$

as atomic number of hydrogen is $1$, nucleus has only one proton

This provides the necessary centripetal force for the electron to remain in orbit.

$\frac{m_e{v}_n^2}{r_n}=\frac{e^2}{4\pi {\epsilon }_o{r}_n^2}..............\left(ii\right)$

Substiuting value of $r_n$ from (ii) in (i), we get:

$m_e{v}_n\frac{e^2}{4\pi {\epsilon }_o{m}_e{v}_n^2}=\frac{nh}{2\pi }$

$v_n=\frac{e^2}{2n{\epsilon }_{o}h}........\left(iii\right)$

Now, kinetic energy of electron is:

$K{E}_n=\frac{1}{2}{m}_e{v}_n^2$

$=\frac{1}{2}{m}_e\frac{e^4}{4n^2{\epsilon }_o^2{h}^2}$

$K{E}_n=\frac{m_e{e}^4}{8n^2{\epsilon }_o^2{h}^2}...........\left(iv\right)$

Potential energy of electron is:

$P{E}_n=-\frac{Q_1{Q}_2}{4\pi {\epsilon }_o{r}_n}$

$=-\frac{e^{2}2\pi m_e{v}_n}{4\pi {\epsilon }_{o}nh}$ using (i)

Substituting $v_n$ from (iii), we get:

$P{E}_n=-\frac{e^2{m}_e}{2{\epsilon }_{o}nh}\times \frac{e^2}{2n{\epsilon }_{o}h}$

$=-\frac{m_e{e}^4}{4n^2{\epsilon }_o^2{h}^2}.............\left(v\right)$

Total energy is hence, given by:

$E_n=K{E}_n+P{E}_n=-\frac{m_e{e}^4}{8n^2{\epsilon }_o^2{h}^2}$

It can be observed that $E_n=-K{E}_n$

Given $E_1=-XeV$

$⟹K{E}_n=XeV$