Hint: For hydrogen atom, force on an
electron due to nucleus is given by,
F = k e 2 r 2 F
= k\dfrac{{{e^2}}}{{{r^2}}} F = k r 2 e 2
Where,
k k k
is constant,
e e e
is charge on electron,
r r r
is distance of electron from nucleus.
Step 1: Note all the given values.
Formula for centripetal force for a particle of mass m m m rotating with velocity v v v in radius r r r is given as,
F = m v 2 r F = \dfrac{{m{v^2}}}{r} F = r m v 2
An electron revolving around a
nucleus of the hydrogen atom. Force on the electron is given by,
F = k e 2 r 2 F
= k\dfrac{{{e^2}}}{{{r^2}}} F = k r 2 e 2
This force acts as centripetal
force on electron responsible for rotation.
Thus,
k e 2 r 2 = m v 2 r k\dfrac{{{e^2}}}{{{r^2}}}
= \dfrac{{m{v^2}}}{r} k r 2 e 2 = r m v 2
⇒ v = k e 2 m r
\Rightarrow v = \sqrt {\dfrac{{k{e^2}}}{{mr}}} ⇒ v = m r k e 2
Step 2: Find an expression for Angular Momentum.
Now, Angular momentum of electron
is given as,
L = m v r L
= mvr L = m v r
⇒ L = m r k e 2 m r
\Rightarrow L = mr\sqrt {\dfrac{{k{e^2}}}{{mr}}} ⇒ L = m r m r k e 2
⇒ L = m r k e 2
\Rightarrow L = \sqrt {mrk{e^2}} ⇒ L = m r k e 2
Since m r k are constants.
⇒ L ∝ r
\Rightarrow L \propto \sqrt r ⇒ L ∝ r
Thus, t he angular momentum of an electron in a hydrogen atom is proportional to the square root of the distance of the electron from the nucleus.