Bohr Model of the Hydrogen Atom

Atoms · CBSE

Theory

What this subtopic is examined on

Stating the three postulates in CBSE's own words (2 marks, set repeatedly).
Deriving $r_n$ — the single most frequently set derivation in the chapter.
Deriving $E_n$ , and deriving $T_n$ (orbital period).
Explaining how Bohr's postulates repair the two failures of Rutherford's model.
Numericals on $r_n\propto n^2$ , on $K=-E$ and $U=2E$ , on excitation and ionisation energy, on the quantum number of a macroscopic orbit.

Why a new model was needed

Rutherford's atom is dynamically like the solar system, but the two differ fundamentally: planets are held by gravity, the electron and nucleus by the Coulomb force between charges.
An object moving in a circle is constantly accelerated (centripetal acceleration), and classical electromagnetic theory says an accelerating charge radiates. So the electron must lose energy continuously, spiral inward and fall into the nucleus — the atom cannot be stable.
As it spirals in, its frequency of revolution changes continuously, and since classically the emitted frequency equals the frequency of revolution, the atom would emit a continuous spectrum. Observation gives a line spectrum.
Classically the ground-state electron ( $v=2.2\times10^6$ m s $^{-1}$ , $r=5.3\times10^{-11}$ m) would radiate at $\nu=\dfrac{v}{2\pi r}=6.6\times10^{15}$ Hz initially, and at a steadily rising frequency after that.
Bohr (1913) concluded that classical electromagnetism, however successful on a large scale, cannot be applied to processes on the atomic scale, and combined classical mechanics with the new quantum ideas of Planck and Einstein in three postulates.

Bohr's three postulates (write them like this)

First postulate — stationary orbits

An electron in an atom can revolve in certain stable orbits without the emission of radiant energy, contrary to the prediction of classical electromagnetic theory.
Each atom therefore has certain definite stable states, each with a definite total energy. These are the stationary states of the atom, and the orbits are called stationary orbits.
This postulate answers the stability problem outright: while it is in a stationary orbit, the electron simply does not radiate.

Second postulate — quantisation of angular momentum

The electron revolves around the nucleus only in those orbits for which the angular momentum is an integral multiple of $h/2\pi$ , where $h$ is Planck's constant.

$L=mv_nr_n=\dfrac{nh}{2\pi},\qquad n=1,2,3,\dots$

$n$ is the principal quantum number. This is the postulate that picks out discrete orbits from the continuum classical physics would allow. (Its physical justification comes from de Broglie — see 12.6.)

Third postulate — the frequency condition

An electron may make a transition from one of its non-radiating orbits to another of lower energy. When it does so, a photon is emitted whose energy equals the difference between the energies of the initial and final states.

$h\nu=E_i-E_f,\qquad E_i>E_f$

Absorption is the same rule reversed: the atom absorbs radiation of exactly the frequency it would emit, and the electron moves to an orbit of higher $n$ : $E_i+h\nu=E_f$ .
This postulate answers the spectrum problem: because $E_i$ and $E_f$ can only take discrete values, only discrete frequencies are radiated — a line spectrum.

Derivation of the radius of the $n$ th orbit (the standard 2- or 3-mark derivation)

In a hydrogen atom, an electron of charge $-e$ and mass $m$ revolves in a circular orbit of radius $r$ about a nucleus of charge $+e$ .

Step 1 — the electrostatic attraction provides the centripetal force:

$\dfrac{mv^2}{r}=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{e^2}{r^2}\qquad\Longrightarrow\qquad mv^2=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{e^2}{r}\qquad\dots(i)$

Step 2 — Bohr's quantisation condition:

$mvr=\dfrac{nh}{2\pi}\qquad\Longrightarrow\qquad v=\dfrac{nh}{2\pi mr}\qquad\dots(ii)$

Step 3 — substitute (ii) into (i):

$m\left(\dfrac{nh}{2\pi mr}\right)^2=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{e^2}{r}\qquad\Longrightarrow\qquad \dfrac{n^2h^2}{4\pi^2mr^2}=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{e^2}{r}$

Step 4 — solve for $r$ :

$r_n=\dfrac{\varepsilon_0n^2h^2}{\pi me^2}\qquad\text{equivalently}\qquad r_n=\dfrac{n^2h^2}{4\pi^2kme^2},\quad k=\dfrac{1}{4\pi\varepsilon_0}$

Hence $r_n\propto n^2$ : the radii of the allowed orbits are in the ratio $1:4:9:16:\dots$
Bohr radius: the radius of the first orbit ( $n=1$ ) of hydrogen,

$a_0=\dfrac{\varepsilon_0h^2}{\pi me^2}=0.529\ \text{\text{Å}}\approx0.53\ \text{\text{Å}}=5.3\times10^{-11}\ \text{m},\qquad r_n=n^2a_0$

For a hydrogen-like atom of nuclear charge $+Ze$ , replace $e^2$ by $Ze^2$ in step 1 and get $r_n=\dfrac{n^2h^2}{4\pi^2mkZe^2}=\dfrac{n^2a_0}{Z}$ .

Derivation of the speed and of the orbital period

Substitute $r_n$ back into (ii):

$v_n=\dfrac{nh}{2\pi mr_n}=\dfrac{e^2}{2\varepsilon_0nh}=\dfrac{2\pi ke^2}{nh}$

So $v_n\propto\dfrac1n$ , with $v_1=2.18\times10^{6}$ m s $^{-1}$ (about $c/137$ ), $v_2=1.09\times10^{6}$ m s $^{-1}$ , $v_3=7.27\times10^{5}$ m s $^{-1}$ .
Orbital period, the standard 2- or 3-mark derivation:

$T_n=\dfrac{2\pi r_n}{v_n}=2\pi\cdot\dfrac{\varepsilon_0n^2h^2}{\pi me^2}\cdot\dfrac{2\varepsilon_0nh}{e^2}=\dfrac{4\varepsilon_0^2n^3h^3}{me^4}$

So $T_n\propto n^3$ . Numerically $T_n=1.52\times10^{-16}\,n^3$ s, giving $T_1=1.5\times10^{-16}$ s, $T_2=1.2\times10^{-15}$ s (the first excited state), $T_3=4.1\times10^{-15}$ s.
Frequency of revolution: $f_n=\dfrac{1}{T_n}=\dfrac{me^4}{4\varepsilon_0^2n^3h^3}\propto\dfrac{1}{n^3}$ .

Derivation of the total energy of the stationary states

Starting again from (i) and (ii), and writing $k=\dfrac{1}{4\pi\varepsilon_0}$ :

Kinetic energy:

$K_n=\dfrac12mv_n^2=\dfrac{ke^2}{2r_n}=\dfrac{ke^2}{2}\times\dfrac{4\pi^2kme^2}{n^2h^2}=\dfrac{2\pi^2k^2me^4}{n^2h^2}$

Potential energy:

$U_n=-\dfrac{ke^2}{r_n}=-ke^2\times\dfrac{4\pi^2kme^2}{n^2h^2}=-\dfrac{4\pi^2k^2me^4}{n^2h^2}$

Total energy:

$E_n=K_n+U_n=\dfrac{2\pi^2k^2me^4}{n^2h^2}-\dfrac{4\pi^2k^2me^4}{n^2h^2}=-\dfrac{2\pi^2k^2me^4}{n^2h^2}=-\dfrac{me^4}{8\varepsilon_0^2n^2h^2}$

Putting in the constants:

$E_n=-\dfrac{2.18\times10^{-18}}{n^2}\ \text{J}=-\dfrac{13.6}{n^2}\ \text{eV}$

The energy is negative because the electron is bound to the nucleus; energy must be supplied to remove it to infinity.
Immediate corollaries, worth writing out every time: $K_n=-E_n=+\dfrac{13.6}{n^2}$ eV and $U_n=2E_n=-\dfrac{27.2}{n^2}$ eV, so $U_n=-2K_n$ .
For hydrogen-like atoms, $E_n=-\dfrac{13.6\,Z^2}{n^2}$ eV.
Since $E_n\propto m$ , replacing the electron by a particle of the same charge but $200$ times the mass multiplies every energy — and the ionisation energy in particular — by $200$ .

12.4.1 Energy levels

The energy is most negative (the atom is most tightly bound) for the orbit nearest the nucleus, $n=1$ . As $n$ increases, $\lvert E_n\rvert$ falls and the energy rises towards zero.
Ground state: $n=1$ , $E_1=-13.6$ eV, radius $a_0$ . At room temperature almost every hydrogen atom is in the ground state.
Excited states: $n>1$ . The atom reaches them by collisions with electrons or other atoms, or by absorbing a photon of exactly the right energy.
Ionisation energy: the minimum energy needed to free the electron from the ground state, $0-(-13.6)=13.6$ eV. Bohr's prediction agrees excellently with the measured value. The binding energy of level $n$ is $\lvert E_n\rvert$ .
Excitation energy from the ground state: $E_n-E_1=13.6\left[1-\dfrac{1}{n^2}\right]$ eV — $10.2$ eV to the first excited state, $12.09$ eV to the second, $12.75$ eV to the third.
As $n$ grows the levels crowd closer together, and the energy needed to free the electron from the excited atom gets smaller.

$n$	$E_n$ (eV)	$K_n$ (eV)	$U_n$ (eV)	$r_n$ (Å)
$1$	$-13.60$	$13.60$	$-27.20$	$0.53$
$2$	$-3.40$	$3.40$	$-6.80$	$2.12$
$3$	$-1.51$	$1.51$	$-3.02$	$4.77$
$4$	$-0.85$	$0.85$	$-1.70$	$8.48$
$\infty$	$0$	$0$	$0$	$\infty$

How Bohr's postulates repair Rutherford's failures

Rutherford's problem	Bohr's repair
An accelerating electron must radiate and spiral in, so the atom collapses in about $10^{-8}$ s	First postulate: certain orbits are stationary — the electron revolves in them without radiating at all
Which orbits? Classically any radius is allowed, so nothing is discrete	Second postulate: only orbits with $L=nh/2\pi$ are allowed, so radii and energies come in a discrete set
A continuously changing frequency gives a continuous spectrum	Third postulate: radiation is emitted only in single quanta on a jump between two discrete levels, $h\nu=E_i-E_f$ , giving discrete lines

Standard results to collect

$L=\dfrac{nh}{2\pi}$ ; for $n=2$ , $L=\dfrac{2h}{2\pi}=\dfrac{h}{\pi}$ .
$r_n=n^2a_0=n^2\times0.53$ Å; $r_3=9\times5.3\times10^{-11}=4.77\times10^{-10}$ m.
$v_n=\dfrac{2.18\times10^6}{n}$ m s $^{-1}$ ; $T_n=1.52\times10^{-16}n^3$ s.
$E_n=-\dfrac{13.6}{n^2}$ eV; $K_n=-E_n$ ; $U_n=2E_n$ .
Two hydrogen atoms in different excited states cannot have the same angular momentum: $E_n$ and $L$ are both fixed by $n$ , so different energies force different $n$ and hence different $L=nh/2\pi$ .
Macroscopic limit: for the earth's orbit ( $m=6.0\times10^{24}$ kg, $v=3\times10^4$ m s $^{-1}$ , $r=1.5\times10^{11}$ m), $n=\dfrac{2\pi mvr}{h}\approx2.6\times10^{74}$ — so large that the quantisation is utterly unnoticeable.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
Using Bohr's atomic model, derive the expression for the radius of $n^{th}$ orbit of the revolving electron in a hydrogen atom. OR Show that the radius of the orbit in hydrogen atom varies as $n^2$ , where $n$ is the principal quantum number of the atom. OR Using Bohr's postulates of the atomic model, derive the expression for the radius of $n^{th}$ electron orbit. Hence obtain the expression for Bohr's radius.
2 marksCBSE Class 12 Physics — SA · 2020 / Delhi 2015 / AI 2014, An · Section B, 2 marksAnswer 1 ↓
2
Write shortcomings of Rutherford atomic model. Explain how these were overcome by the postulates of Bohr's atomic model.
2 marksCBSE Class 12 Physics — SA · 2020, U · Section B, 2 marksAnswer 2 ↓
3
State Bohr's quantization condition of angular momentum. Calculate the shortest wavelength of the Brackett series and state to which part of the electromagnetic spectrum does it belong.
2 marksCBSE Class 12 Physics — SA · Delhi 2019 · Section B, 2 marksAnswer 3 ↓
4
Find out the wavelength of the electron orbiting in the ground state of hydrogen atom.
2 marksCBSE Class 12 Physics — SA · Delhi 2017, U · Section B, 2 marksAnswer 4 ↓
5
State Bohr postulate of hydrogen atom that gives the relationship for the frequency of emitted photon in a transition.
2 marksCBSE Class 12 Physics — SA · Foreign 2016 · Section B, 2 marksAnswer 5 ↓
6
(a) How did de Broglie hypothesis provide an explanation for Bohr's second postulate for quantisation of orbital angular momentum of the orbiting electron in hydrogen atom? Discuss. (b) Identify the transition of electron in Bohr model which gives rise to (i) the maximum, and (ii) the minimum wavelength in the Balmer series of hydrogen spectrum.
3 marksCBSE Class 12 Physics — SA · 2022C · Section C, 3 marksAnswer 6 ↓
7
Using Bohr's postulates, derive the expression for the orbital period of the electron moving in the $n^{th}$ orbit of hydrogen atom.
3 marksCBSE Class 12 Physics — SA · AI 2019, U · Section C, 3 marksAnswer 7 ↓
8
A hydrogen atom initially in the ground state absorbs a photon which excites it to the $n=4$ level. Estimate the frequency of the photon.
3 marksCBSE Class 12 Physics — SA · 2018 · Section C, 3 marksAnswer 8 ↓
9
(a) The radius of the innermost electron orbit of a hydrogen atom is $5.3\times10^{-11}$ m. Calculate its radius in $n=3$ orbit. (b) The total energy of an electron in the first excited state of the hydrogen atom is $-3.4$ eV. Find out its (i) kinetic energy, and (ii) potential energy in this state.
3 marksCBSE Class 12 Physics — SA · Delhi 2014C, Ev · Section C, 3 marksAnswer 9 ↓
10
(a) Write two important limitations of Rutherford model which could not explain the observed features of atomic spectra. How were these explained in Bohr's model of hydrogen atom? (b) Using Bohr's postulates, obtain the expression for the radius of the $n^{th}$ orbit in hydrogen atom.
5 marksCBSE Class 12 Physics — Long Answer · 4/5, Delhi 2015C, An · Section E, 5 marksAnswer 10 ↓
11
Using Bohr's postulates, derive the expression for the total energy of the electron in the stationary states of the hydrogen atom.
5 marksCBSE Class 12 Physics — Long Answer · 3/5, Foreign 2014 · Section E, 5 marksAnswer 11 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

12
The radius of the $n^{th}$ orbit in Bohr model of hydrogen atom is proportional to
- (a)
  $n^2$
- (b)
  $\dfrac1{n^2}$
- (c)
  $n$
- (d)
  $\dfrac1n$
1 markCBSE Class 12 Physics — Multiple Choice · 2023, U · Section A, 1 markAnswer 12 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
In a hydrogen atom, an electron of charge $-e$ revolves around the nucleus of charge $+e$ in a circular orbit of radius $r$ , such that the necessary centripetal force is provided by the electrostatic force of attraction between the electron and nucleus: $\;\dfrac{mv^2}r=\dfrac1{4\pi\varepsilon_0}\cdot\dfrac{e^2}{r^2}\;\Rightarrow\;mv^2=\dfrac1{4\pi\varepsilon_0}\cdot\dfrac{e^2}r\quad\dots(i)$ From Bohr's quantization condition: $\;mvr=\dfrac{nh}{2\pi}\;\Rightarrow\;v=\dfrac{nh}{2\pi mr}\quad\dots(ii)$ Substituting (ii) into (i): $\;m\left(\dfrac{nh}{2\pi mr}\right)^2=\dfrac1{4\pi\varepsilon_0}\cdot\dfrac{e^2}r\;\Rightarrow\;\dfrac{n^2h^2}{4\pi^2mr^2}=\dfrac1{4\pi\varepsilon_0}\cdot\dfrac{e^2}r$ $\;\Rightarrow\;r=\dfrac{n^2h^2\varepsilon_0}{\pi me^2}$ , where $n=1,2,3,\dots$ is the principal quantum number. This gives the radius of the $n^{th}$ orbit of the H-atom — the radii of the orbits increase proportionally with $n^2$ , i.e. $r\propto n^2$ . The radius of the first orbit ( $n=1$ ) of the H-atom is called the Bohr radius $a_0$ : $\;a_0=\dfrac{h^2\varepsilon_0}{\pi me^2}=0.529\text{ \text{Å}}$ , so the radius of the $n^{th}$ orbit becomes $\;r=n^2\times0.529\text{ \text{Å}}$ .
2.↑ back to the question
Rutherford's atomic model is inconsistent with classical physics: according to electromagnetic theory, an electron is a charged particle moving in a circular orbit around the nucleus and is accelerated, so it should emit radiation continuously and thereby lose energy; consequently the radius of the electron's orbit would decrease continuously, the atom should produce a continuous spectrum, and ultimately the electron would fall into the nucleus, so the atom would collapse (in about $10^{-8}$ s). But atoms are fairly stable and emit a discrete LINE spectrum, not a continuous one. Bohr's postulates resolved these observed features: (i) Quantum (quantization) condition — of all the possible circular orbits allowed by classical theory, electrons are permitted to circulate only in those orbits in which the angular momentum of an electron is an integral multiple of $h/2\pi$ : $\;L=mvr=\dfrac{nh}{2\pi}$ . (ii) Stationary orbits — while revolving in the permissible orbits, an electron does NOT radiate energy; these non-radiating orbits are called stationary orbits, resolving the stability problem. (iii) Frequency condition — an atom can emit or absorb radiation in the form of discrete energy photons only when an electron jumps from a higher to a lower orbit (or vice versa), giving a discrete spectrum: $\;h\nu=E_i-E_f$ .
3.↑ back to the question
Bohr's quantization condition: the electron can revolve around the nucleus only in those circular orbits for which the angular momentum is an integral multiple of $h/2\pi$ , i.e. $\;mvr=\dfrac{nh}{2\pi},\;n=1,2,3,\dots$ The shortest wavelength (series limit) of the Brackett series ( $n_1=4$ , $n_2=\infty$ ) is given by: $\;\dfrac1\lambda=1.097\times10^7\left[\dfrac1{4^2}-\dfrac1{\infty^2}\right]=\dfrac{1.097\times10^7}{16}$ $\;\lambda=1.4585\times10^{-6}\text{ m}$ This wavelength lies in the INFRARED region of the electromagnetic spectrum.
4.↑ back to the question
(Interpreting this as the de Broglie wavelength associated with the emission when the electron is in the ground state, using the ground state energy): $\;\lambda=\dfrac{hc}E=\dfrac{6.62\times10^{-34}\times3\times10^8}{-13.6\text{ eV}}$ (taking the magnitude of the ground state energy in Joules, $13.6\times1.6\times10^{-19}$ J): $\;\lambda=\dfrac{6.62\times10^{-34}\times3\times10^8}{13.6\times1.6\times10^{-19}}=\dfrac{1.986\times10^{-25}}{21.76\times10^{-19}}\approx9.126\times10^{-8}\text{ m}=912.6\text{ \text{Å}}$
5.↑ back to the question
Frequency condition of the Bohr model: an atom can emit or absorb radiation in the form of discrete energy photons only when an electron jumps from a higher to a lower orbit or from a lower to a higher orbit, respectively: $\;h\nu=E_i-E_f$ where $\nu$ is the frequency of radiation emitted (or absorbed), and $E_i$ and $E_f$ are the energies associated with the stationary orbits of principal quantum numbers $n_i$ and $n_f$ respectively (where $n_i>n_f$ for emission).
6.↑ back to the question
(a) According to de Broglie, the electron in its circular orbit, as proposed by Bohr, must be seen as a particle wave. When a string is plucked, a vast number of wavelengths are excited, but only those wavelengths that have nodes at the ends survive and form the standing wave in the string. Similarly, according to de Broglie's hypothesis, a stationary orbit is one that contains an integral number of de Broglie waves associated with the revolving electron. The total distance covered by the electron equals the circumference of the orbit, i.e. $\;2\pi r_n=n\lambda,\;n=1,2,3,\dots\quad\dots(i)$ Now, according to the de Broglie wavelength, $\;\lambda=\dfrac h{mv}$ . Putting this into equation (i): $\;2\pi r_n=\dfrac{nh}{mv_n}\;\Rightarrow\;mv_nr_n=\dfrac{nh}{2\pi}$ , which is exactly Bohr's quantum condition for the angular momentum of the electron. (b) For the Balmer series, transitions terminate at $n_1=2$ . (i) The MAXIMUM wavelength (smallest energy gap) corresponds to the transition from $n_2=3$ to $n_1=2$ (the first line of the series). (ii) The MINIMUM wavelength (series limit) corresponds to the transition from $n_2=\infty$ to $n_1=2$ .
7.↑ back to the question
When the electron revolves around the nucleus in a certain specified circular orbit without the emission of radiant energy, centripetal force $=$ electrostatic force: $\;\dfrac{mv_n^2}{r_n}=\dfrac1{4\pi\varepsilon_0}\cdot\dfrac{e^2}{r_n^2}\quad\dots(i)$ From Bohr's postulate of angular momentum: $\;mv_nr_n=\dfrac{nh}{2\pi}\quad\dots(ii)$ From (i) and (ii): $\;r_n=\dfrac{n^2h^2\varepsilon_0}{\pi me^2}$ and $\;v_n=\dfrac{e^2}{2\varepsilon_0nh}$ Time period of revolution of electron in the $n^{th}$ orbit: $\;T_n=\dfrac{2\pi r_n}{v_n}=\dfrac{4\varepsilon_0^2n^3h^3}{me^4}$
8.↑ back to the question
$\Delta E=E_4-E_1=-\dfrac{13.6}{16}-(-13.6)=13.6\left(1-\dfrac1{16}\right)=13.6\times\dfrac{15}{16}$ eV $\;h\nu=13.6\times\dfrac{15}{16}\times1.6\times10^{-19}\text{ J}$ $\;\nu=\dfrac{13.6\times\dfrac{15}{16}\times1.6\times10^{-19}}{6.63\times10^{-34}}\approx3\times10^{15}\text{ Hz}$
9.↑ back to the question
(a) $\;r_n\propto n^2\;\Rightarrow\;\dfrac{r_3}{r_1}=3^2=9\;\Rightarrow\;r_3=9r_1=9\times5.3\times10^{-11}=4.77\times10^{-10}\text{ m}$ (b) (i) Kinetic energy, $E_k=-E=-(-3.4)=3.4$ eV (ii) Potential energy, $E_p=2E=2\times(-3.4)=-6.8$ eV
10.↑ back to the question
(a) Rutherford's model could not explain (i) the stability of matter — an accelerating electron should radiate energy continuously, causing it to spiral into the nucleus and the atom to collapse, contrary to observation; (ii) the characteristic (discrete) line spectrum of atoms — classical theory predicts a continuous spectrum, whereas atoms emit sharp, discrete spectral lines. Bohr's model resolved these by postulating (i) stationary (non-radiating) orbits, in which the electron does not lose energy despite being in circular motion, resolving the stability problem; and (ii) the frequency condition $h\nu=E_i-E_f$ , whereby radiation is emitted only in discrete quanta corresponding to transitions between specific, quantized energy levels — naturally producing a discrete line spectrum rather than a continuous one. (b) Same derivation as Q16 above: centripetal force = electrostatic force gives $mv^2=\dfrac1{4\pi\varepsilon_0}\cdot\dfrac{e^2}r$ , and Bohr's quantization condition gives $v=\dfrac{nh}{2\pi mr}$ ; combining these, $\;r=\dfrac{n^2h^2\varepsilon_0}{\pi me^2}$ , the radius of the $n^{th}$ orbit of the hydrogen atom.
11.↑ back to the question
According to Bohr's postulates, in a hydrogen atom a single electron revolves around a nucleus of charge $+e$ . For an electron moving with uniform speed in a circular orbit of a given radius, the centripetal force is provided by the Coulomb force of attraction between the electron and the nucleus (gravitational attraction being negligible): $\;\dfrac{mv^2}r=\dfrac{ke^2}{r^2}\;\left(k=\dfrac1{4\pi\varepsilon_0}\right)\;\Rightarrow\;mv^2=\dfrac{ke^2}r\quad\dots(i)$ By Bohr's second postulate: $\;mvr=\dfrac{nh}{2\pi}\;\Rightarrow\;v=\dfrac{nh}{2\pi mr}$ . Substituting into (i): $\;r=\dfrac{n^2h^2}{4\pi^2kme^2}\quad\dots(ii)$ Kinetic energy: $\;E_k=\dfrac12mv^2=\dfrac{ke^2}{2r}=\dfrac{ke^2}2\times\dfrac{4\pi^2kme^2}{n^2h^2}=\dfrac{2\pi^2k^2me^4}{n^2h^2}$ Potential energy: $\;E_p=-\dfrac{ke^2}r=-ke^2\times\dfrac{4\pi^2kme^2}{n^2h^2}=-\dfrac{4\pi^2k^2me^4}{n^2h^2}$ Total energy: $\;E=E_k+E_p=\dfrac{2\pi^2k^2me^4}{n^2h^2}-\dfrac{4\pi^2k^2me^4}{n^2h^2}=-\dfrac{2\pi^2k^2me^4}{n^2h^2}=-\dfrac{13.6}{n^2}\text{ eV}$
12.Answer: (A)↑ back to the question
Radius, $r=n^2r_0$ , i.e. $r\propto n^2$ . Option (a).