Question

Using Rutherford model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?
OR
Using Bohr's postulates of the atomic model derive the expression for radius of nth electron orbit. Hence obtain the expression for Bohrs radius.

Answer 1

According to Rutherford's model, we have

$\frac{m{v}^2}{r}=\frac{1}{4\pi {ϵ}_0}\frac{Z{e}^2}{r^2}$

$⟹$ $=\frac{1}{4\pi {ϵ}_0}\frac{Z{e}^2}{r}$

Therefore,

Total Energy $=P.E+K.E$

$T.E=-\frac{1}{4\pi {ϵ}_0}\frac{Z{e}^2}{r}+\frac{1}{2}m{v}^2$

$=\frac{1}{2}\frac{1}{4\pi {ϵ}_0}\frac{Z{e}^2}{r}$

$=-\frac{1}{8\pi {ϵ}_0}\frac{Z{e}^2}{r}$

Energy is negative implies that the electron- nucleus is a bound or attractive system.

OR

According to Bohr's atomic model, electrons revolve around the nucleus only in those orbits for which the angular momentum is an integral multiple of $\frac{h}{2\pi }$

So, as per Bohr's postulate, we have

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

Therefore,

$\frac{m{v}^2}{r}=\frac{1}{4\pi {ϵ}_0}\frac{Z{e}^2}{r^2}$

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

Therefore,

$m^2{v}^2{r}^2=\frac{n^2{h}^2}{4{\pi }^2}$

This implies,

$r=\frac{{ϵ}_0{n}^2{h}^2}{\pi Z{e}^{2}m}$

This is the required expression for Bohr's radius.