Question

The radius of an orbit of a hydrogen atom is $0.85nm$. Calculate the velocity (in m/sec) of the electron in its orbit.

Answer 1

We need to know that Niels Bohr was a physicist in Danish, he received a nobel prize for his work of understanding quantum theory and atomic structure. Bohr developed a Bohr model of an atom, an electron energy levels are discrete and that the electrons revolve the atomic nucleus, in stable orbits, but electrons jump from one orbit to another, as energy levels.

Complete step by step answer:
We must know that a simple planar representation of an electron is called orbit. The line segments from its centre to its perimeter is called a radius, of a circle or sphere. Radius comes from Latin word which means ray, but ray also means the spoke of a chariot wheel.
Given the value, the radius of an orbit of a hydrogen atom is $0.85nm$ .
According to the one of the postulates of Bohr’s theory (an electron revolves in circular orbits and it is restricted by angular momentum, the angular momentum of an electron is an integral multiple of
$h/2\pi$ ),
$mvr = \dfrac{{nh}}{{2\pi }}$ -----$(1)$
Where,
$m =$ mass of an electron$= 9.1 \times {10^{ - 31}}kg$
$h =$ Planck’s constant$= 6.626 \times {10^{34}}Js$
$n =$ principal quantum number of orbit
($n$ is calculated from,$R = {R_0} \times \dfrac{{{n^2}}}{{{Z^2}}}$
Where, $R =$ radius and $Z =$ atomic number, ${R_0} = 0.529 \times {10^{ - 1}}$
Substitute in above equation,
$0.85 = 0.529 \times {10^{ - 1}} \times \dfrac{{{n^2}}}{1}$
${n^2} = 16$
$n = 4$
Rewrite the above equation $(1)$,
$\nu = \dfrac{{nh}}{{2\pi (mr)}}$
Radius of orbit in hydrogen is calculated from the below formula,
${r_n} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}km{e^2}}}$ ----$(2)$
Substituting equation$(2)$in question $(1)$,
$v = \dfrac{{nh \times 4\pi \times km{e^2}}}{{2\pi (m{n^2}{h^2})}}$
On simplification we get,
$v = \dfrac{{2\pi k{e^2}}}{{nh}}$
Where, $v$ is velocity (in m/sec), $k$ is Coulomb’s constant $= 9 \times {10^9}Jm/{C^2}$ and $e =$ charge of an electron $= 1.6 \times {10^{ - 19}}C$ .
Substitute the values in above equation,
$v = \dfrac{{2 \times 3.14(9 \times {{10}^9}){{(1.6 \times {{10}^{ - 19}})}^2}}}{{(6.626 \times {{10}^{ - 34}}) \times 4}}$
On simplification we get,
$v = \dfrac{{144.69 \times {{10}^{ - 29}}}}{{26.05 \times {{10}^{ - 34}}\;}}$
On dividing we get,
$v = 5.55 \times {10^5}$
The velocity of an electron in its orbit is $v = 5.55 \times {10^5}m/\sec$ .

Note:
We need to remember that where two charged points orbit each other at less than that of light speeds in a system the Bohr model of atom gives almost exact results. The Bohr model not only involves hydrogen (one electron system), ionized helium and lithium, but it also involves Rydberg states and positronium, where the electron is far away.