Question

Consider the hydrogen atom to be a proton embedded in a cavity of radius $a_0$ (Bohr's radius), whose charge is neutralized by the addition of an electron to the cavity in vacuum, infinitely slowly. Then the wavelength of the electron when it is at a distance of $a_0$ from the proton will be

• A. $\lambda =\sqrt{\frac{4\pi {\epsilon }_0{a}_0}{h^2{e}^{2}m}}$
• B. $\lambda =\sqrt{\frac{4\pi {\epsilon }_0{a}_0{h}^2}{em}}$
• C. $\lambda =\sqrt{\frac{4\pi {\epsilon }_0{a}_0{h}^2}{e^{2}m}}$
• D. $\lambda =\sqrt{\frac{4\pi {\epsilon }_0{a}_0{h}^2}{e^{5}m}}$

Answer 1

Answer: (C)

$\lambda =\sqrt{\frac{4\pi {\epsilon }_0{a}_0{h}^2}{e^{2}m}}$

Work obtained in the neutralization process is given by :

$W=-{\int }_{\infty }^{a_0}F\cdot da=-{\int }_{\infty }^{a_0}\frac{1}{4\pi {\epsilon }_0}\frac{(-)e^2}{a_0^2}\cdot d{a}_0$

$W=-\frac{e^2}{4\pi {\epsilon }_0\cdot a_0}$

This work is to be called as potential energy. However in doing so, one should note that this energy is simply lost during the process of attraction between proton and electron. As reported in the problem the electron is brought infinitely slowly, with this condition the electron simply possesses only potential energy. Thus,

$TE=PE+KE=PE=-\frac{e^2}{4\pi {\epsilon }_0\cdot a_0}$ .........(i)
Now in order, the electron to be captured by the proton to form a ground state hydrogen atom, it should also attain kinetic energy $e^2/\left(8\pi {\epsilon }_0{a}_0\right)$ (as it is half of the potential energy given in the question). Thus, the total energy of the electron if it attains the ground state in $H$-atom,
$TE=PE+KE=-\frac{e^2}{4\pi {\epsilon }_0{a}_0}+\frac{e^2}{8\pi {\epsilon }_0{a}_0}$
$TE=-\frac{e^2}{8\pi {\epsilon }_0{a}_0}$
The wavelength of electron when it is simply at a distance $a_0$ from the proton can be given as :

$\lambda =\frac{h}{mv}=hp$

Also, $KE=\frac{1}{2}m{v}^2=\frac{p^2}{2m}$ $(\because p=mv)$

Thus, $\lambda =\frac{h}{\sqrt{2m\left(KE\right)}}$
Since, $KE=0$ at this situation, thus $\lambda =\infty$
Also, when electron is at a distance $a_0$ in Bohr's orbit of $H-$ atom.

$\lambda =\frac{h}{\sqrt{2m\left(KE\right)}}=\frac{h}{\sqrt{\frac{2m{e}^2}{2a_0\cdot 4\pi {\epsilon }_0}}}$

$\lambda =\sqrt{\frac{4\pi {\epsilon }_0{a}_0{h}^2}{e^{2}m}}$