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) $$\lambda = \sqrt {\dfrac{{4\pi {\varepsilon _0}{a_0}}}{{{h^2}{e^2}m}}} $$
(B) $$\lambda = \sqrt {\dfrac{{4\pi {\varepsilon _0}{a_0}{h^2}}}{{em}}} $$
(C) $$\lambda = \sqrt {\dfrac{{4\pi {\varepsilon _0}{a_0}{h^2}}}{{{e^2}m}}} $$
(D) $$\lambda = \sqrt {\dfrac{{4\pi {\varepsilon _0}{a_0}{h^2}}}{{{e^5}m}}} $$

Answer 1

Hint:Bohr’s model was proposed by Neil Bohr in 1915. It was the modification of the Rutherford model of the atom. He stated that in the atomic structure the electrons are able to move in fixed orbitals called shells. He further explained the five energy levels of each shell or orbital.

Complete step-by-step answer:Work which is obtained in the neutralization process can be shown by the following formula:
$$W = - \int_\infty ^{{a_0}} {F.da = \int_\infty ^{{a_0}} {\dfrac{1}{{4\pi {\varepsilon _0}}}} } \times \dfrac{{( - ){e^2}}}{{{a_0}^2}}.d{a_0}$$
On simplification we get
$$W = - \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}{a_0}^{}}}$$
The above calculated work is the potential energy. As we know that the energy is lost during the attraction between proton and electron. As it is given in the question that the electron is brought infinitely slowly, so this means in this condition on potential energy will be possessing. So total energy will be
$$TE = PE + KE = PE = - \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}{a_0}^{}}}$$
Now as the electron is to be captured by protons to form a ground state of hydrogen atom. It will have the kinetic energy equal to half of potential energy. So the total energy will be
$$TE = PE + KE = PE = - \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}{a_0}^{}}} + \dfrac{{{e^2}}}{{8\pi {\varepsilon _0}{a_0}^{}}}$$
$$TE = - \dfrac{{{e^2}}}{{8\pi {\varepsilon _0}{a_0}^{}}}$$
The wavelength of the electron will be
$$\lambda = \dfrac{h}{{mv}} = hp(p = mv)$$
Here p is the momentum.
So the formula for kinetic energy is
$$KE = \dfrac{1}{2}m{v^2} = \dfrac{{{p^2}}}{{2m}}$$
So the wavelength will be
$$\lambda = \dfrac{h}{{\sqrt {2m(KE)} }}$$
In the given situation the kinetic energy will be zero and wavelength is at infinity. So the electron at the distance$${a_0}$$ in Bohr’s orbit of hydrogen atom will be
$$\lambda = \dfrac{h}{{\sqrt {2m(KE)} }} = \dfrac{h}{{\sqrt {\dfrac{{2m{e^2}}}{{2{a_0}.4\pi {\varepsilon _0}}}} }} = \sqrt {\dfrac{{4\pi {\varepsilon _0}{a_0}{h^2}}}{{{e^2}m}}} $$

So the correct answer is option (C).

Note: The postulate of Bohr’s model was that the electron revolves around the positively charged nucleus. The orbits have fixed energy levels. The electrons in the orbit move from lower energy level to higher energy level. The Bohr’s model was violating the Heisenberg uncertainty principle. It was not able to explain the reason for obtaining spectra from larger atoms.