Question

Consider the hydrogen atom to be a proton embedded in a cavity of radius a (Bohr radius) whose charge is neutralized by the addition of an electron to the cavity in vacuum, infinitely slowly.
If the magnitude of average kinetic energy is half the magnitude of the average potential energy, then which of the following options is/ are correct.

• A. Potential energy of hydrogen atom is equal to $-\frac{e^2}{4\pi {\epsilon }_0{a}_0}$
• B. The magnitude of total energy is equal to the magnitude of kinetic energy.
• C. The total energy is one fourth of the potential energy.
• D. Work done in neutralization is equal to $-\frac{e^2}{4\pi {\epsilon }_0{a}_0}$

Answer 1

Answer: (A), (B), (D)

Work done in neutralization is equal to $-\frac{e^2}{4\pi {\epsilon }_0{a}_0}$

When charge on the proton is neutralized, work is done.
Let work done = W
Work done (W) = Force x distance (to bring the charge from infinite to given point)
$W=-\underset{\infty }{\overset{a_0}{\int }}\frac{e^2}{4\pi {\epsilon }_0{a}_0^2}d{a}_0=-\frac{e^2}{4\pi {\epsilon }_0{a}_0}$

$\text{where}{\epsilon }_0=\text{permittivity}$

The work is equal to potential energy as velocity becomes zero or it has no K.E
$K.E=0$.
$P.E=\frac{-e^2}{4\pi {\epsilon }_0{a}_0}$
Since the proton captures the electron and forms a ground state in a hydrogen atom, the electron possesses most K.E which is generally half of the potential energy.
$K.E=\frac{e^2}{8\pi {\epsilon }_0{a}_0}$
Thus total energy $(T.E)=P.E+K.ET.E=\frac{-e^2}{4\pi {\epsilon }_0{a}_0}+\frac{e^2}{8\pi {\epsilon }_0{a}_0}=\frac{-e^2}{8\pi {\epsilon }_0{a}_0}$

which is equivalent to kinetic energy but with a opposite sign.