Chapter 12: Atoms

In Rutherford's nuclear model the electron's orbit (radius $$\approx 10^{-10}\,\mathrm{m}$$) is much larger than the nucleus (radius $$\approx 10^{-15}\,\mathrm{m}$$). The ratio of these two sizes is

$$\frac{r_{\text{orbit}}}{r_{\text{nucleus}}} = \frac{10^{-10}\,\mathrm{m}}{10^{-15}\,\mathrm{m}} = 10^{5}$$

If the solar system were built to the same proportions, the radius of the earth's orbit would have to be $$10^{5}$$ times the radius of the sun:

$$r_{\text{orbit}}' = 10^{5} \times r_{\text{sun}} = 10^{5} \times (7 \times 10^{8}\,\mathrm{m}) = 7 \times 10^{13}\,\mathrm{m}$$

The actual radius of the earth's orbit is $$1.5 \times 10^{11}\,\mathrm{m}$$. Comparing the two,

$$\frac{r_{\text{orbit}}'}{r_{\text{actual}}} = \frac{7 \times 10^{13}\,\mathrm{m}}{1.5 \times 10^{11}\,\mathrm{m}} \approx 4.7 \times 10^{2}$$

The scaled orbit would be about $$467$$ times bigger than the real one. Hence, if the dimensions of the solar system had the same proportions as those of the atom, the earth would be very much farther away from the sun than it actually is. In other words, the atom is far more 'empty' relative to its nucleus than the solar system is relative to the sun.

At the distance of closest approach $$d$$ the $$\alpha$$-particle is momentarily at rest, so by conservation of energy all of its initial kinetic energy $$K$$ has been converted into the electrostatic potential energy of the $$\alpha$$-particle and the nucleus.

An $$\alpha$$-particle carries charge $$+2e$$ and the gold nucleus carries charge $$+Ze$$ with $$Z = 79$$. Equating kinetic energy to potential energy at closest approach:

$$K = \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{d}$$

Solving for the distance of closest approach:

$$d = \frac{1}{4\pi\varepsilon_0}\,\frac{2Ze^2}{K}$$

Convert the kinetic energy to joules:

$$K = 7.7\,\mathrm{MeV} = 7.7 \times 10^{6} \times 1.6 \times 10^{-19}\,\mathrm{J} = 1.232 \times 10^{-12}\,\mathrm{J}$$

Substituting $$\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\,\mathrm{N\,m^2\,C^{-2}}$$, $$Z = 79$$ and $$e = 1.6 \times 10^{-19}\,\mathrm{C}$$:

$$d = \frac{(9 \times 10^{9})(2)(79)(1.6 \times 10^{-19})^2}{1.232 \times 10^{-12}}$$

$$d = \frac{3.64 \times 10^{-26}}{1.232 \times 10^{-12}} \approx 3.0 \times 10^{-14}\,\mathrm{m}$$

The distance of closest approach is about $$3.0 \times 10^{-14}\,\mathrm{m}$$, i.e. $$30\,\mathrm{fm}$$. This sets an upper limit on the size of the gold nucleus.

The energy required to pull the atom apart into a free proton and a free electron is the ionisation energy, which equals the magnitude of the total energy of the bound electron. Hence the total energy of the electron is

$$E = -13.6\,\mathrm{eV} = -13.6 \times 1.6 \times 10^{-19}\,\mathrm{J} = -2.18 \times 10^{-18}\,\mathrm{J}$$

Orbital radius. For the electron moving in a circle of radius $$r$$, the Coulomb attraction provides the centripetal force:

$$\frac{1}{4\pi\varepsilon_0}\,\frac{e^2}{r^2} = \frac{mv^2}{r} \quad\Rightarrow\quad mv^2 = \frac{1}{4\pi\varepsilon_0}\,\frac{e^2}{r}$$

So the kinetic and potential energies are

$$K = \tfrac12 mv^2 = \frac{1}{8\pi\varepsilon_0}\,\frac{e^2}{r}, \qquad U = -\frac{1}{4\pi\varepsilon_0}\,\frac{e^2}{r}$$

and the total energy is

$$E = K + U = -\frac{1}{8\pi\varepsilon_0}\,\frac{e^2}{r}$$

Solving for $$r$$:

$$r = -\frac{1}{4\pi\varepsilon_0}\,\frac{e^2}{2E} = -\frac{(9 \times 10^{9})(1.6 \times 10^{-19})^2}{2 \times (-2.18 \times 10^{-18})}$$

$$r = \frac{2.30 \times 10^{-28}}{4.36 \times 10^{-18}} \approx 5.3 \times 10^{-11}\,\mathrm{m}$$

Orbital speed. The kinetic energy equals the magnitude of the total energy, $$K = -E = 2.18 \times 10^{-18}\,\mathrm{J}$$. From $$K = \tfrac12 mv^2$$ with $$m = 9.1 \times 10^{-31}\,\mathrm{kg}$$:

$$v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 2.18 \times 10^{-18}}{9.1 \times 10^{-31}}} = \sqrt{4.79 \times 10^{12}}$$

$$v \approx 2.2 \times 10^{6}\,\mathrm{m/s}$$

According to classical electromagnetic theory, an electron moving in a circular orbit is continuously accelerating, and an accelerated charge radiates electromagnetic waves. The frequency of the radiated wave equals the frequency of revolution of the electron, i.e. the number of revolutions it completes per second:

$$\nu = \frac{v}{2\pi r}$$

where $$v$$ is the orbital speed and $$2\pi r$$ is the circumference of the orbit. Using the values found in Example 12.3, $$v = 2.2 \times 10^{6}\,\mathrm{m/s}$$ and $$r = 5.3 \times 10^{-11}\,\mathrm{m}$$:

$$\nu = \frac{2.2 \times 10^{6}}{2\pi \times 5.3 \times 10^{-11}} = \frac{2.2 \times 10^{6}}{3.33 \times 10^{-10}}$$

$$\nu \approx 6.6 \times 10^{15}\,\mathrm{Hz}$$

So the electron would initially emit light of frequency about $$6.6 \times 10^{15}\,\mathrm{Hz}$$, which lies in the ultraviolet region. But as it radiates, the electron loses energy, its orbit shrinks, and its speed and frequency keep changing. Classically the atom would therefore emit a continuous range of frequencies while the electron spirals into the nucleus — contradicting both the observed sharp line spectra and the stability of atoms. This failure of classical theory is what Bohr's model was introduced to resolve.

Given : The original Rutherford experiment used a $$2\,\mathrm{MeV\; –\; 8\,MeV}$$ beam of $$\alpha$$–particles (charge $$+2e$$, mass $$m_{\alpha}=4u$$) and a thin gold foil whose nuclei have charge $$Z_{\mathrm{Au}}=79$$ and mass $$M_{\mathrm{Au}}\simeq 197u$$.

We now replace the gold foil by a thin sheet of solid hydrogen. Each scatterer is a proton (hydrogen nucleus) of charge $$Z_{\mathrm H}=1$$ and mass $$M_{\mathrm H}=m_{p}=1u.$$ The kinetic energy of the incident $$\alpha$$–particles is assumed to be the same as in the gold–foil experiment.

1. Effect of the nuclear charge on the scattering probability

The differential (Rutherford) cross–section for scattering through an angle $$\theta$$ is

$$\frac{\mathrm d \sigma}{\mathrm d\Omega}=\left(\frac{1}{4\pi\varepsilon_{0}}\,\frac{2Ze^{2}}{4E\sin ^{2}(\theta/2)}\right)^{2}\;.$$

For the same $$\alpha$$–energy $$E$$, the ratio of the hydrogen cross–section to the gold cross–section is

$$\frac{\left( Z_{\mathrm H}e^{2}\right)^{2}}{\left( Z_{\mathrm{Au}}e^{2}\right)^{2}}=\left(\frac{1}{79}\right)^{2}\approx 1.6\times10^{-4}. $$

Hence the probability of an $$\alpha$$–particle being scattered through a given large angle is smaller by a factor of about $$6\,000$$. Almost every $$\alpha$$ passes straight through.

2. Limitation on the largest possible laboratory angle

Besides the small cross–section there is a kinematic restriction because the target (proton) is much lighter than the projectile (alpha).

For an elastic collision between a projectile of mass $$m$$ and a target of mass $$M$$ initially at rest, the laboratory scattering angle $$\theta_{\text{lab}}$$ is related to the centre–of–mass (CM) scattering angle $$\theta_{\text{CM}}$$ by

$$\tan \theta_{\text{lab}}=\frac{\sin \theta_{\text{CM}}}{\cos \theta_{\text{CM}}+m/M}. $$

With $$m=m_{\alpha}=4u$$ and $$M=m_{p}=1u$$ we get

$$\tan \theta_{\text{lab}}=\frac{\sin \theta_{\text{CM}}}{\cos \theta_{\text{CM}}+4}. $$

The numerator never exceeds 1 while the denominator is always > 3, so

$$\theta_{\text{lab(max)}}=\arctan\left(\frac{1}{4}\right)\approx 14^{\circ}. $$

Therefore an $$\alpha$$–particle cannot be scattered through angles larger than about $$14^{\circ}$$ when the target is a proton. Back–scattering ( $$\theta \gtrsim 90^{\circ}$$ ) is impossible.

3. Recoil of target nuclei

The energy transferred to a proton in a head-on elastic collision is

$$E_{p}^{\;\text{max}}=E_{\alpha}\,\frac{4Mm}{(M+m)^{2}}=E_{\alpha}\,\frac{4(1)(4)}{(1+4)^{2}}=\tfrac{16}{25}\,E_{\alpha}\approx0.64E_{\alpha}. $$

Thus a fast recoil proton can indeed be produced, but it emerges at a fairly large angle while the $$\alpha$$ itself continues almost in its original direction.

4. Expected experimental observations

• Almost the entire incident beam reaches the detector placed straight ahead; its count rate is practically the same as with no foil present.
• No detectable $$\alpha$$–particles appear at angles beyond about $$15^{\circ}$$, so the famous back-scattering of Rutherford would be absent.
• The foil emits energetic recoil protons (not present with a gold target). Detecting them would confirm momentum transfer, but that was not part of Rutherford’s original arrangement.

Conclusion

Replacing the gold by solid hydrogen eliminates large-angle scattering of $$\alpha$$–particles: almost all of them pass through undeflected, while some fast recoil protons may be observed.

Given

• Energy difference between the two levels: $$\Delta E = 2.3\,\mathrm{eV}$$
• 1 electron-volt: $$1\,\mathrm{eV} = 1.602\times10^{-19}\,\mathrm{J}$$
• Planck's constant: $$h = 6.626\times10^{-34}\,\mathrm{J\,s}$$

Step 1: Convert $$\Delta E$$ to joule

$$\Delta E = 2.3\,\mathrm{eV}\times1.602\times10^{-19}\,\mathrm{J\,eV^{-1}}$$

$$\Delta E = 3.6846\times10^{-19}\,\mathrm{J}$$

Step 2: Use $$\Delta E = h\nu$$ to find the frequency $$\nu$$

$$\nu = \dfrac{\Delta E}{h} = \dfrac{3.6846\times10^{-19}\,\mathrm{J}}{6.626\times10^{-34}\,\mathrm{J\,s}}$$

$$\nu = 0.556\times10^{15}\,\mathrm{Hz} = 5.6\times10^{14}\,\mathrm{Hz}$$

Therefore, the radiation emitted during the transition has a frequency of approximately $$5.6\times10^{14}\,\mathrm{Hz}$$.

For an electron in a stationary (Bohr) orbit of a hydrogen atom the force is purely coulombic, i.e. $$F \propto 1/r^{2}$$ (so the potential energy $$U \propto -1/r$$). For such an inverse-square law force the virial theorem gives a fixed relation between the time-averaged kinetic energy $$K$$ and potential energy $$U$$:

$$U = -2K$$

The total (mechanical) energy of the electron is therefore

$$E = K + U = K - 2K = -K$$

or, equivalently,

$$K = -E \quad\text{and}\quad U = 2E$$

In the ground state of hydrogen the given total energy is

$$E = -13.6\,\mathrm{eV}$$

Hence

Kinetic energy

$$K = -E = -(-13.6\,\mathrm{eV}) = 13.6\,\mathrm{eV}$$

Potential energy

$$U = 2E = 2\times(-13.6\,\mathrm{eV}) = -27.2\,\mathrm{eV}$$

Given data

• Ground state (initial level): $$n_i = 1$$
• Excited state (final level): $$n_f = 4$$
• Hydrogen‐atom energy levels: $$E_n = -\dfrac{13.6\;\text{eV}}{n^2}$$
• Planck constant: $$h = 6.626\times10^{-34}\;\text{J s}$$
• Speed of light: $$c = 3.00\times10^8\;\text{m s}^{-1}$$
• Conversion: $$1\;\text{eV} = 1.602\times10^{-19}\;\text{J}$$

1. Energies of the two levels

Ground state:

$$E_1 = -\dfrac{13.6\,\text{eV}}{1^2} = -13.6\;\text{eV}$$

Fourth state:

$$E_4 = -\dfrac{13.6\,\text{eV}}{4^2} = -\dfrac{13.6}{16}\;\text{eV} = -0.85\;\text{eV}$$

2. Energy absorbed by the atom

$$\Delta E = E_4 - E_1 = (-0.85) - (-13.6)\;\text{eV} = 12.75\;\text{eV}$$

3. Convert this energy to joules

$$\Delta E = 12.75\;\text{eV}\times1.602\times10^{-19}\;\dfrac{\text{J}}{\text{eV}} = 2.04255\times10^{-18}\;\text{J}$$

4. Wavelength of the photon

Photon energy is related to wavelength by $$E = \dfrac{hc}{\lambda}$$, so

$$\lambda = \dfrac{hc}{\Delta E}$$

Substituting,

$$\lambda = \dfrac{(6.626\times10^{-34}\;\text{J s})(3.00\times10^8\;\text{m s}^{-1})}{2.04255\times10^{-18}\;\text{J}}$$

$$\lambda = 9.73\times10^{-8}\;\text{m} = 97.3\;\text{nm}$$

5. Frequency of the photon

Using $$\nu = \dfrac{c}{\lambda}$$:

$$\nu = \dfrac{3.00\times10^8\;\text{m s}^{-1}}{9.73\times10^{-8}\;\text{m}} = 3.08\times10^{15}\;\text{Hz}$$

Result

The photon that excites the hydrogen atom from $$n = 1$$ to $$n = 4$$ has

$$\lambda \;\approx\; 97.3\;\text{nm}, \qquad \nu \;\approx\; 3.08\times10^{15}\;\text{Hz}. $$

Given data

Radius of the first Bohr orbit (ground state, $$n = 1$$):
$$r_1 = 5.3 \times 10^{-11}\,\mathrm{m}$$

Bohr’s radius formula

For a hydrogen atom the radius of the $$n^{\text{th}}$$ stationary orbit is
$$r_n = n^{2} a_0$$
where $$a_0$$ is the Bohr radius. Since the given $$r_1$$ itself equals $$a_0$$, we have

$$a_0 = r_1 = 5.3 \times 10^{-11}\,\mathrm{m}$$

(i) Radius for $$n = 2$$

$$\begin{aligned} r_2 &= 2^{2} a_0 \\ &= 4 \times 5.3 \times 10^{-11}\,\mathrm{m} \\ &= 2.12 \times 10^{-10}\,\mathrm{m} \end{aligned}$$

(ii) Radius for $$n = 3$$

$$\begin{aligned} r_3 &= 3^{2} a_0 \\ &= 9 \times 5.3 \times 10^{-11}\,\mathrm{m} \\ &= 4.77 \times 10^{-10}\,\mathrm{m} \end{aligned}$$

Thus the second and third Bohr orbit radii are four and nine times the first-orbit radius, respectively.

The stationary-state energies of hydrogen are given by

$$E_n = -\dfrac{13.6}{n^{2}}\;\mathrm{eV}\qquad (n = 1,2,3,\ldots)$$

To know how high an atom can be excited we equate the largest possible energy gain to the incident electron energy $$12.5\,\mathrm{eV}$$:

$$\Delta E_{1\to n} = E_n - E_1 = -\frac{13.6}{n^{2}} - (-13.6) = 13.6\Bigl(1 - \frac{1}{n^{2}}\Bigr) \le 12.5$$

$$1 - \frac{1}{n^{2}} \le \frac{12.5}{13.6} = 0.9191 \;\;\Rightarrow\;\; \frac{1}{n^{2}} \ge 0.0809$$

$$n^{2} \le \frac{1}{0.0809} \approx 12.4 \;\;\Longrightarrow\;\; n_{\max} = 3$$

Hence collisions can raise atoms at most to the level $$n = 3$$. The jump $$1\to 3$$ needs $$12.09\,\mathrm{eV}$$ (less than the available $$12.5\,\mathrm{eV}$$), whereas $$1\to 4$$ would need $$12.75\,\mathrm{eV}$$ (more than $$12.5\,\mathrm{eV}$$).

Once an atom reaches $$n = 3$$ it can return to lower levels through all permissible downward transitions:

• $$3 \to 2$$
• $$3 \to 1$$
• $$2 \to 1$$ (after the intermediate population of level 2)

The wavelength of each emitted photon follows the Rydberg relation

$$\frac{1}{\lambda} = R\Bigl(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\Bigr),\qquad R = 1.097\times10^{7}\;\mathrm{m^{-1}}$$

Thus only three spectral lines appear: one in the visible (Balmer) region and two in the ultraviolet (Lyman) region.

Given data

• Radius of the Earth’s orbit: $$r = 1.5 \times 10^{11}\,\mathrm{m}$$
• Orbital speed of the Earth: $$v = 3.0 \times 10^{4}\,\mathrm{m\,s^{-1}}$$
• Mass of the Earth: $$m = 6.0 \times 10^{24}\,\mathrm{kg}$$
• Planck’s reduced constant: $$\hbar = \frac{h}{2\pi} = 1.055 \times 10^{-34}\,\mathrm{J\,s}$$

Bohr’s quantisation condition

For a particle moving in a circular orbit, Bohr postulated

$$m v r = n\hbar$$

where $$n$$ is the (principal) quantum number.

Step 1: Calculate the Earth’s orbital angular momentum

$$ L = m v r = (6.0 \times 10^{24})(3.0 \times 10^{4})(1.5 \times 10^{11}) $$

First multiply the mantissas:

$$6.0 \times 3.0 = 18.0; \quad 18.0 \times 1.5 = 27.0$$

Next add the exponents of ten:

$$10^{24} \times 10^{4} \times 10^{11} = 10^{24+4+11} = 10^{39}$$

Because we obtained 27.0 (not between 1 and 10) we write it as $$2.7 \times 10^{1}$$ and add one more power of ten:

$$L = 2.7 \times 10^{1} \times 10^{39}\;\mathrm{kg\,m^{2}\,s^{-1}} = 2.7 \times 10^{40}\,\mathrm{kg\,m^{2}\,s^{-1}}$$

Step 2: Solve for the quantum number $$n$$

$$ n = \frac{L}{\hbar} = \frac{2.7 \times 10^{40}}{1.055 \times 10^{-34}} $$

Separate the mantissas and the powers of ten:

$$ \frac{2.7}{1.055} \approx 2.56, \quad 10^{40} \div 10^{-34} = 10^{40-(-34)} = 10^{74} $$

$$ \therefore \; n \approx 2.56 \times 10^{74} $$

Rounded to two significant figures,

$$n \approx 2.6 \times 10^{74}$$

Conclusion

The principal quantum number that would characterise the Earth’s revolution around the Sun, if Bohr’s quantisation applied, is of the order of $$10^{74}$$ — an astronomically large number.