JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 8)
The acceleration of an electron in the first orbit of the hydrogen atom (n = 1) is :
$${{{h^2}} \over {{\pi ^2}{m^2}{r^3}}}$$
$${{{h^2}} \over {{8\pi ^2}{m^2}{r^3}}}$$
$${{{h^2}} \over {{4\pi ^2}{m^2}{r^3}}}$$
$${{{h^2}} \over {{4\pi }{m^2}{r^3}}}$$
설명
The speedy of the particle in the orbit of an atom is $$v = {{{I^2}} \over {2h{\varepsilon _0}}}$$
We have the radius of the first orbit is
$$r = {{{h^2}{\varepsilon _0}} \over {\pi m{e^2}}}$$
$$ \Rightarrow {\varepsilon _0} = {{r\pi m{e^2}} \over {{h^2}}}$$
Therefore, the acceleration of an electron in the first orbit of the hydrogen atom (n = 1) is
$${{{v^2}} \over r} = {{{I^6}\pi m} \over {4{h^2}\varepsilon _0^3}} = {{{h^2}} \over {4{r^3}{\pi ^2}{m^2}}}$$