Bohr Model — Radius, Velocity & Energy · Atoms and Nuclei

PHYAtoms — Bohr Model & Spectra

Bohr Model — Radius, Velocity & Energy

Bohr's model pictures the electron in a hydrogen-like atom moving in stationary circular orbits whose angular momentum is quantised in units of $\hbar = h/2\pi$ . From this one postulate fall out the orbit radius $r_n \propto n^2/Z$ , the orbital speed $v_n \propto Z/n$ , and the total energy $E_n \propto -Z^2/n^2$ , with the ground state of hydrogen pinned at $-13.6\,\text{eV}$ and $a_0 = 0.53\,\text{Å}$ .

ISC examines this constantly because, once you fix the $n=1$ values for hydrogen, every other orbit and every hydrogen-like ion ( $\text{He}^+, \text{Li}^{2+}$ ) follows by simple scaling — no need to re-derive constants.

Radius of the n-th orbit (hydrogen-like)

r_n = \dfrac{n^2}{Z}\,a_0, \qquad a_0 = 0.53\,\text{Å}

r_n

is the orbit radius,

n

the principal quantum number,

Z

the atomic number,

a_0 = 0.53\,\text{Å} = 0.53\times 10^{-10}\,\text{m}

the Bohr radius (hydrogen ground state). For hydrogen (

Z=1

r_2 = 4a_0

; for

\text{Li}^{2+}

(

Z=3

), the first orbit is

a_0/3

Orbital speed of the electron

v_n = \dfrac{Z}{n}\,v_1, \qquad v_1 = 2.188\times 10^{6}\ \text{m s}^{-1}

v_n

is the speed in the

n

-th orbit,

v_1 = 2.188\times 10^{6}\,\text{m s}^{-1}

the speed in the first Bohr orbit of hydrogen (

\approx c/137

, the fine-structure ratio). Speed falls as

1/n

, so

v_2 = v_1/2

for hydrogen.

Total energy of the n-th orbit

E_n = -13.6\,\dfrac{Z^2}{n^2}\ \text{eV}

E_n

is the total (bound) energy, negative because the electron is trapped;

13.6\,\text{eV}

is the hydrogen ground-state binding energy. For hydrogen

E_2 = -3.4\,\text{eV}

and

E_3 = -1.51\,\text{eV}

; energy rises toward

0

n

increases.

Kinetic, potential and total energy relations

K = -E_n = +13.6\,\dfrac{Z^2}{n^2}\ \text{eV}, \qquad U = 2E_n = -2K

K

is the kinetic energy,

U

the electrostatic potential energy,

E_n

the total energy. The virial theorem gives

E_n = -K

and

U = 2E_n

; for hydrogen's ground state

K = +13.6\,\text{eV}

U = -27.2\,\text{eV}

, summing to

E_1 = -13.6\,\text{eV}

Bohr quantisation of angular momentum

m v_n r_n = n\,\dfrac{h}{2\pi}

m

is the electron mass,

h

Planck's constant,

n = 1, 2, 3, \dots

the orbit number. This postulate selects the allowed (stationary) orbits; the electron in such an orbit does not radiate.

Treat the $n=1$ hydrogen values ( $a_0 = 0.53\,\text{Å}$ , $v_1 = 2.188\times 10^6\,\text{m s}^{-1}$ , $E_1 = -13.6\,\text{eV}$ ) as anchors, then scale: $r \propto n^2/Z$ , $v \propto Z/n$ , $E \propto Z^2/n^2$ .
For a hydrogen-like ion of charge $Z$ , orbits shrink by $Z$ and bind more tightly by $Z^2$ — that is why $\text{Li}^{2+}$ ( $Z=3$ ) has a tiny first orbit ( $0.53/3 \approx 0.177\,\text{Å}$ ) and a deep $-122.4\,\text{eV}$ ground state.
The energies are negative because the electron is bound; the magnitude is the energy needed to free it (the ionisation energy from that level).
Kinetic and total energy have equal magnitude but opposite sign: $K = |E_n|$ . The potential energy is twice the total energy and is negative.
Speed in the first orbit is about $c/137$ , comfortably non-relativistic, which is why the simple Bohr formulae work for the hydrogen atom.
Bohr's model is exact only for one-electron systems (H, $\text{He}^+$ , $\text{Li}^{2+}$ ); it cannot handle multi-electron atoms because of electron-electron repulsion.
Energy difference between two levels is $E_{n_2} - E_{n_1}$ using signed values; the photon energy emitted or absorbed equals this gap (covered under the photon subtopic).

Where the marks go

Forgetting the $Z$ factors when the atom is not hydrogen: $\text{Li}^{2+}$ needs $r = n^2 a_0/Z$ with $Z=3$ , not the bare hydrogen radius.
Sign slips on energy: total and potential energies are negative, kinetic energy is positive. Writing $K = E_n$ (negative) instead of $K = -E_n$ is a common error.
Scaling speed as $1/n^2$ instead of $1/n$ . Only radius goes as $n^2$ and energy as $1/n^2$ ; speed goes as $Z/n$ .
Mixing units of length — leaving $a_0$ in Å while the rest of the working is in metres ( $1\,\text{Å} = 10^{-10}\,\text{m}$ ).

How the board asks it

Numericalthe radius, speed and energy scaling $r \propto n^2/Z$ , $v \propto Z/n$ , $E \propto -Z^2/n^2$
The radius of the first Bohr orbit of hydrogen is $0.53\,\text{Å}$ and the ground-state energy is $-13.6\,\text{eV}$ . Calculate the radius of the second orbit and the energy of the electron in the third orbit of the hydrogen atom.
Derive / provebohr quantisation of angular momentum and the coulomb-centripetal balance
Using Bohr's postulate of quantisation of angular momentum, derive an expression for the radius of the $n$ -th orbit of the electron in a hydrogen atom in terms of $n$ , $h$ , $m$ , $e$ and $\varepsilon_0$ .
Applicationthe $Z$ -scaling for hydrogen-like ions ( $\text{He}^+$ , $\text{Li}^{2+}$ )
For the $\text{Li}^{2+}$ ion ( $Z = 3$ ), calculate the radius of the first orbit and the ground-state energy, given $a_0 = 0.53\,\text{Å}$ and $E_1 = -13.6\,\text{eV}$ for hydrogen.
Numericalthe kinetic, potential and total energy relations $K = -E_n$ , $U = 2E_n$
The total energy of the electron in the ground state of hydrogen is $-13.6\,\text{eV}$ . Find its kinetic energy and its potential energy in this orbit.
Define / stateenergies are negative because the electron is bound; ionisation energy equals $|E_n|$
Why is the total energy of an electron in any Bohr orbit of the hydrogen atom negative? State the physical significance of this negative sign.

Practise this topic

NextIonisation, Excitation & Spectral Lines

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.