Question

In a hydrogen atom, the electron and proton are bound at a distance of about $0.53\stackrel{o}{A}$:

A) Estimate the potential energy of the system in $eV$, taking the zero of the potential energy at infinite separation of the electron from proton.

B) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?

C) What are the answers to (a) and (b) above if the zero of potential energy is taken at $1.06\stackrel{o}{A}$ separtion?

Answer 1

A) Potential energy of the system = Potential energy at given separation potential energy at infinite separation
$U=\frac{k{q}_1{q}_2}{r}$
Putting all the values
$U=\frac{9\times {10}^9\times 1.6\times {10}^{-19}\times -1.6\times {10}^{-19}}{0.53\times {10}^{-10}}$
$U=-43.47\times {10}^{-19}J$

Potential energy in $eV$ :-
As $1eV=1.6\times {10}^{-19}J$

$U=\frac{-43.47\times {10}^{-19}eV}{1.6\times {10}^{-19}}\Rightarrow U=-27.16eV$


B)Given,
Kinetic energy of the $e^{-}=\frac{1}{2}\left|Potentialenergyofthesystem\right|$
$=\frac{1}{2}\times 27.16=13.58eV$
Totel energy of $e^{-}$ = Kinetic energy of the $e^{-}$ $+$ Potential energy of the system
$=13.58-27.16$
$=-13.58eV$
To make free $e^{-}$ free $T.E>0$
So, externally $e^{-}$ required $13.58eV$ energy which is also called as $"lonizationenergy"$.


C) Potential energy of the sytem = Potential energy at given separation$-$Potential energy at $1.06\stackrel{o}{A}$ separation ...(i)
We calculated the Potential energy at given separtion i.e., $-27.16eV$
Potential energy at $1.06\stackrel{o}{A}$ separation,
$U=\frac{k{q}_1{q}_2}{r}$
$U=\frac{9\times {10}^9\times 1.6\times {10}^{-19}\times -1.6\times {10}^{-19}}{1.06\times {10}^{-10}}$
$U=-21.73\times {10}^{-19}J$

Potential energy in $eV$
$U=-13.58eV$

Putting values in equation ...(i)
Potential energy of the system $=-27.16-(-13.58)=-13.58eV$
Kinetic energy of the $e^{-}$ = 13.58 $eV$
Total energy of $e^{-}$ = Kinetic energy of the $e^{-}$ $+$ Potential energy of the system
$=13.58eV-13.58eV=0$
Total energy of $e^{-}=0eV$
lonization energy required to move the $e^{-}$ to infinity is 13.58 $eV$.