Question

In the Bohr model of the hydrogen atom, let \[R\] , \[v\] and \[E\] represent the radius of the orbit, speed of the electron and the total energy of the electron respectively. Which of the following quantities are proportional to the quantum number \[n\] ?
A. \[vR\]
B. \[RE \\ \]
C. $\dfrac{v}{E} \\ $ ​
D. $\dfrac{R}{E}$

Answer 1

Hint: To answer this question, we find the proportionality of the principal quantum number \[n\] with radius of orbit, speed of electron and the total energy of electron. After that we will check each option given in the question to find which quantity is proportional to the principal quantum number \[n\] .

Formula Used:
Radius of orbit,
$R = \left( {\dfrac{{{h^2}{ \in _o}}}{{\pi m{e^2}}}} \right){n^2}$
Speed of electron,
$v = \left( {\dfrac{{{e^2}}}{{2{ \in _o}h}}} \right)\dfrac{1}{n}$
Energy of electron,
$E = - \left( {\dfrac{{m{e^4}}}{{8{ \in _o}{h^2}}}} \right)\dfrac{1}{{{n^2}}}$
Where $h$ is Planck’s constant, ${ \in _o}$ is permittivity of free space, $e$ is the charge on a fundamental electron, $m$ is the magnetic quantum number and $n$ is the principal quantum number.

Complete step by step solution:
The electrons in the Bohr model of the atom circle the nucleus in well-defined circular orbits. The quantum number n is used to identify the orbits by an integer. By releasing or absorbing energy, electrons can move from one orbit to another.

We know that, $R = \left( {\dfrac{{{h^2}{ \in _o}}}{{\pi m{e^2}}}} \right){n^2}$ which implies that, $R \propto {n^2}$ ...(1)
(Since all other quantities are constant)
Similarly, $v = \left( {\dfrac{{{e^2}}}{{2{ \in _o}h}}} \right)\dfrac{1}{n}$ which implies that $v \propto \dfrac{1}{n}$ ...(2)
And $E = - \left( {\dfrac{{m{e^4}}}{{8{ \in _o}{h^2}}}} \right)\dfrac{1}{{{n^2}}}$ which implies that $E \propto \dfrac{1}{{{n^2}}}$ ...(3)

Now, we will check each option given in the question one-by-one using equation (1), (2) and (3).
From option A.,
we have $vR = \dfrac{1}{n} \times {n^2} = n$ which is proportional to $n$ . ...(4)
From option B.,
we have $RE = {n^2} \times \dfrac{1}{{{n^2}}}$ which is a constant and thus not proportional to $n$ .
From option C.,
$\dfrac{v}{E} = \dfrac{{\dfrac{1}{n}}}{{\dfrac{1}{{{n^2}}}}}$ which gives, $\dfrac{v}{E} = n$ which is proportional to $n$.
From option D.,
$\dfrac{R}{E} = \dfrac{{{n^2}}}{{\dfrac{1}{{{n^2}}}}}$ which gives, $\dfrac{R}{E} = {n^4}$ which is not proportional to $n$ . ...(5)
Thus, we see that \[vR\] and $\dfrac{v}{E}$ are proportional to $n$ .

Hence, option A and C are answers.

Note: The electrons revolve around the nucleus in circular orbits. The radius around the orbit has some special values of radius. In these stationary orbits they do not radiate energy as expected from Maxwell’s laws. Energy of each stationary orbits are fixed, electrons can jump from a higher orbit to lower orbit by emitting a photon of radiation