Question

What should be the kinetic energy of electron in the first orbit of hydrogen atom?

Answer 1

According to Bohr's postulate-

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

Squaring both sides, we get

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

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

Dividing the above equation by $2$, we get

$\frac{1}{2}m{v}^2=\frac{n^2{h}^2}{8{\pi }^{2}m{r}^2}.....\left(1\right)$

As we know that,

$K.E.=\frac{1}{2}m{v}^2.....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$K.E.=\frac{n^2{h}^2}{8{\pi }^{2}m{r}^2}$

As we know that,

$r=0.529\frac{n^2}{Z}\mathring{A}=0.529\times {10}^{-10}m\frac{n^2}{Z}$

$n=1\left(\text{Given}\right)$

$Z=1$

$\therefore r=0.529\times {10}^{-10}\frac{1^2}{1}=0.529\times {10}^{-10}m$

Molar mass of hydrogen atom, $\left(m\right)=1$

$\therefore K.E.=\frac{1^2\times 6.62\times 6.62\times {10}^{-68}}{8\times 3.14\times 3.14\times 0.529\times 0.529\times {10}^{-20}\times 1}$

$\Rightarrow K.E.=1.98\times {10}^{-48}J=1.23\times {10}^{-29}eV$