Question
The acceleration of an electron in the first orbit of the hydrogen atom $$\left( {z = 1} \right)$$ is :
A.
$$\frac{{{h^2}}}{{{\pi ^2}{m^2}{r^3}}}$$
B.
$$\frac{{{h^2}}}{{8{\pi ^2}{m^2}{r^3}}}$$
C.
$$\frac{{{h^2}}}{{4{\pi ^2}{m^2}{r^3}}}$$
D.
$$\frac{{{h^2}}}{{4\pi {m^2}{r^3}}}$$
Answer :
$$\frac{{{h^2}}}{{4{\pi ^2}{m^2}{r^3}}}$$
Solution :
Speed of electron in first orbit $$\left( {n = 1} \right)$$ of hydrogen atom $$\left( {z = 1} \right),$$
$$\eqalign{
& v = \frac{{{e^2}}}{{2{\varepsilon _0}h}} \cr
& r = \frac{{{h^2}{\varepsilon _0}}}{{\pi m{e^2}}} \Rightarrow {\varepsilon _0} = \frac{{r\pi m{e^2}}}{{{h^2}}} \cr} $$
Acceleration of electron,
$$\eqalign{
& \frac{{{v^2}}}{r} = \frac{{{e^4}}}{{4\varepsilon _0^2{h^2}}} \times \frac{{\pi m{e^2}}}{{{h^2}{\varepsilon _0}}} \cr
& = \frac{{{e^4} \times \pi m{e^2}}}{{4{h^4}\varepsilon _0^2}}\,......\left( {{\text{ii}}} \right) \cr} $$
eliminating $${\varepsilon _0}$$
$$ = \frac{{{e^4}\pi m{e^2}{h^6}}}{{4{h^4}{r^3}{\pi ^3}{m^3}{e^6}}} = \frac{{{h^2}}}{{4{\pi ^2}{m^2}{r^3}}}$$