Atoms

(a) No different from.

In both Thomson's model and Rutherford's model, the overall size of the atom is of the same order (~$10^{-10}$ m). The difference lies in the internal structure, not the overall atomic size.

(b) Thomson's model / Rutherford's model.

In Thomson's model, electrons are embedded in a uniform positive charge distribution. At the equilibrium positions the net electrostatic force on each electron is zero, so they are in stable equilibrium. In Rutherford's model, the electrons revolve around the nucleus and are always under the influence of the Coulomb attractive force (centripetal force), so they always experience a net force.

(c) Rutherford's model.

In Rutherford's model, electrons revolve in circular orbits and undergo centripetal acceleration. An accelerating charge radiates energy (according to classical electrodynamics), so the electron would continuously lose energy and spiral into the nucleus — the atom would collapse.

(d) Thomson's model / Rutherford's model.

In Thomson's model, the positive charge is spread uniformly throughout the atom, giving a nearly continuous (uniform) mass distribution. In Rutherford's model, almost all the mass is concentrated in the tiny nucleus, giving a highly non-uniform mass distribution.

(e) Both the models.

In Thomson's model the positive charge (and hence most of the mass) is distributed throughout the sphere. In Rutherford's model the positive charge and most of the mass are concentrated in the nucleus. In both models, the positively charged part possesses most of the mass (since electron mass is negligible compared to nuclear/atomic mass).

Given: Alpha-particle scattering experiment is repeated with a thin sheet of solid hydrogen instead of gold foil.

Key facts:
- Mass of alpha particle ($m_\alpha$) $= 4$ u
- Mass of hydrogen nucleus (proton, $m_p$) $= 1$ u
- Since m_\alpha > m_p, the alpha particle is heavier than the target nucleus.

Analysis:

In the original Rutherford experiment with gold ($A = 197$ u), the gold nucleus is much heavier than the alpha particle, so the gold nucleus remains essentially stationary and the alpha particle can be scattered back.

When an alpha particle (mass $4$ u) collides with a hydrogen nucleus (proton, mass $1$ u), by the laws of elastic collision, the lighter target (proton) will be knocked forward with a speed up to twice the alpha particle's speed, while the alpha particle itself will continue forward with reduced speed.

Using conservation of momentum and energy for a head-on elastic collision:
$$v'_\alpha = \frac{m_\alpha - m_p}{m_\alpha + m_p}\,v_0 = \frac{4-1}{4+1}\,v_0 = \frac{3}{5}\,v_0$$

The alpha particle cannot be scattered backward (large-angle scattering) because it is heavier than the proton. The maximum scattering angle for the alpha particle is less than $90°$.

Expected Result: No large-angle (backward) scattering of alpha particles will be observed. The alpha particles will mostly pass through with small deflections, and the protons (hydrogen nuclei) will be knocked forward. This experiment would not reveal the nuclear structure in the way Rutherford's gold-foil experiment did.

Given:
$$\Delta E = E_i - E_f = 2.3 \text{ eV} = 2.3 \times 1.6 \times 10^{-19} \text{ J} = 3.68 \times 10^{-19} \text{ J}$$

Formula used (Bohr's frequency condition):
$$h\nu = \Delta E$$
$$\nu = \frac{\Delta E}{h}$$

Calculation:
$$\nu = \frac{3.68 \times 10^{-19}}{6.626 \times 10^{-34}}$$
$$\boxed{\nu \approx 5.55 \times 10^{14} \text{ Hz}}$$

This frequency lies in the visible region of the electromagnetic spectrum.

Given: Total energy in ground state, $E_1 = -13.6$ eV

Concept: For an electron in a circular orbit, the electrostatic force provides the centripetal force:
$$\frac{e^2}{4\pi\varepsilon_0 r^2} = \frac{mv^2}{r}$$

This gives:
$$\frac{1}{2}mv^2 = \frac{e^2}{8\pi\varepsilon_0 r} \quad \Rightarrow \quad K = -\frac{1}{2}\,U$$

where $U = -\dfrac{e^2}{4\pi\varepsilon_0 r}$ is the potential energy.

Also, total energy:
$$E = K + U = -\frac{e^2}{8\pi\varepsilon_0 r} = \frac{U}{2} = -K$$

Therefore:
$$K = -E_1 = -(-13.6) = +13.6 \text{ eV}$$
$$U = 2E_1 = 2 \times (-13.6) = -27.2 \text{ eV}$$

Results:
- Kinetic energy $= +13.6$ eV
- Potential energy $= -27.2$ eV

*Verification:* $E = K + U = 13.6 + (-27.2) = -13.6$ eV ✓

Given:
- Initial state: $n_i = 1$ (ground level), $E_1 = -13.6$ eV
- Final state: $n_f = 4$

Energy of the $n = 4$ level:
$$E_4 = \frac{-13.6}{4^2} = \frac{-13.6}{16} = -0.85 \text{ eV}$$

Energy of the absorbed photon:
$$\Delta E = E_4 - E_1 = -0.85 - (-13.6) = 12.75 \text{ eV}$$
$$\Delta E = 12.75 \times 1.6 \times 10^{-19} = 2.04 \times 10^{-18} \text{ J}$$

Frequency of the photon:
$$\nu = \frac{\Delta E}{h} = \frac{2.04 \times 10^{-18}}{6.626 \times 10^{-34}}$$
$$\boxed{\nu \approx 3.08 \times 10^{15} \text{ Hz}}$$

Wavelength of the photon:
$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{3.08 \times 10^{15}}$$
$$\boxed{\lambda \approx 9.74 \times 10^{-8} \text{ m} = 97.4 \text{ nm}}$$

This wavelength lies in the ultraviolet region (Lyman series).

Formula for speed in Bohr's model:
$$v_n = \frac{e^2}{4\pi\varepsilon_0}\cdot\frac{1}{n\hbar} = \frac{v_1}{n}$$

where $v_1 = \dfrac{e^2}{4\pi\varepsilon_0 \hbar} \approx 2.18 \times 10^6$ m/s

(a) Speeds:

$$v_1 = \frac{2.18 \times 10^6}{1} = 2.18 \times 10^6 \text{ m/s}$$

$$v_2 = \frac{2.18 \times 10^6}{2} = 1.09 \times 10^6 \text{ m/s}$$

$$v_3 = \frac{2.18 \times 10^6}{3} = 7.27 \times 10^5 \text{ m/s}$$

(b) Orbital period:

The radius of the $n$-th orbit: $r_n = n^2 a_0$, where $a_0 = 5.3 \times 10^{-11}$ m

Orbital period:
$$T_n = \frac{2\pi r_n}{v_n} = \frac{2\pi n^2 a_0}{v_1/n} = \frac{2\pi n^3 a_0}{v_1}$$

$$T_1 = \frac{2\pi \times (5.3 \times 10^{-11})}{2.18 \times 10^6} = \frac{3.33 \times 10^{-10}}{2.18 \times 10^6} \approx 1.527 \times 10^{-16} \text{ s}$$

$$\boxed{T_1 \approx 1.53 \times 10^{-16} \text{ s}}$$

$$T_2 = T_1 \times 2^3 = 1.53 \times 10^{-16} \times 8 \approx 1.22 \times 10^{-15} \text{ s}$$

$$\boxed{T_2 \approx 1.22 \times 10^{-15} \text{ s}}$$

$$T_3 = T_1 \times 3^3 = 1.53 \times 10^{-16} \times 27 \approx 4.13 \times 10^{-15} \text{ s}$$

$$\boxed{T_3 \approx 4.13 \times 10^{-15} \text{ s}}$$

Given: $r_1 = a_0 = 5.3 \times 10^{-11}$ m

Formula (Bohr's model):
$$r_n = n^2 a_0$$

For $n = 2$:
$$r_2 = (2)^2 \times 5.3 \times 10^{-11} = 4 \times 5.3 \times 10^{-11}$$
$$\boxed{r_2 = 2.12 \times 10^{-10} \text{ m}}$$

For $n = 3$:
$$r_3 = (3)^2 \times 5.3 \times 10^{-11} = 9 \times 5.3 \times 10^{-11}$$
$$\boxed{r_3 = 4.77 \times 10^{-10} \text{ m}}$$

Given: Energy of electron beam = 12.5 eV; hydrogen is at room temperature (ground state, $n = 1$, $E_1 = -13.6$ eV).

Finding the highest excited level reached:

Energy of level $n$: $E_n = -13.6/n^2$ eV

The electron can excite hydrogen from ground state ($n=1$) to level $n$ if:
$$E_n - E_1 \leq 12.5 \text{ eV}$$
$$-\frac{13.6}{n^2} - (-13.6) \leq 12.5$$
$$13.6\left(1 - \frac{1}{n^2}\right) \leq 12.5$$
$$1 - \frac{1}{n^2} \leq 0.919$$
$$\frac{1}{n^2} \geq 0.081 \Rightarrow n^2 \leq 12.3 \Rightarrow n \leq 3.5$$

So the maximum level reached is $n = 3$.

Check: $E_3 - E_1 = -13.6/9 - (-13.6) = 13.6(1 - 1/9) = 13.6 \times 8/9 = 12.09$ eV < 12.5 eV ✓

Check $n=4$: $E_4 - E_1 = 13.6(1-1/16) = 12.75$ eV > 12.5 eV ✗

Possible transitions from $n = 3$:

The electrons can be in $n = 1, 2, 3$. The possible downward transitions are:

1. $n = 3 \to n = 1$: $\Delta E = 12.09$ eV → Lyman series (UV)
2. $n = 3 \to n = 2$: $\Delta E = -1.51 - (-3.4) = 1.89$ eV → Balmer series (visible)
3. $n = 2 \to n = 1$: $\Delta E = -3.4 - (-13.6) = 10.2$ eV → Lyman series (UV)

Wavelengths emitted:

Using $\lambda = hc/\Delta E$:

- $n=3\to n=1$: $\lambda = \dfrac{1240}{12.09} \approx 102.6$ nm (Lyman series)
- $n=2\to n=1$: $\lambda = \dfrac{1240}{10.2} \approx 121.6$ nm (Lyman series)
- $n=3\to n=2$: $\lambda = \dfrac{1240}{1.89} \approx 656.6$ nm (Balmer series)

Conclusion: The emitted spectrum will contain wavelengths belonging to the Lyman series (ultraviolet) and the Balmer series (visible). Specifically, three spectral lines at approximately 102.6 nm, 121.6 nm (Lyman), and 656.6 nm (Balmer) will be emitted.

Given:
- Orbital radius: $r = 1.5 \times 10^{11}$ m
- Orbital speed: $v = 3 \times 10^4$ m/s
- Mass of earth: $m = 6.0 \times 10^{24}$ kg
- $h = 6.626 \times 10^{-34}$ J·s

Bohr's quantisation condition:
$$L = mvr = \frac{nh}{2\pi}$$

Solving for $n$:
$$n = \frac{mvr \cdot 2\pi}{h} = \frac{2\pi m v r}{h}$$

Substituting values:
$$n = \frac{2\pi \times 6.0 \times 10^{24} \times 3 \times 10^4 \times 1.5 \times 10^{11}}{6.626 \times 10^{-34}}$$

Numerator:
$$2\pi \times 6.0 \times 10^{24} \times 3 \times 10^4 \times 1.5 \times 10^{11}$$
$$= 2\pi \times 2.7 \times 10^{40}$$
$$= 6.2832 \times 2.7 \times 10^{40}$$
$$= 1.6965 \times 10^{41}$$

$$n = \frac{1.6965 \times 10^{41}}{6.626 \times 10^{-34}}$$

$$\boxed{n \approx 2.56 \times 10^{74}}$$

Conclusion: The quantum number $n \approx 2.56 \times 10^{74}$, which is an enormously large number. This is consistent with the correspondence principle — at very large quantum numbers, quantum mechanics merges with classical mechanics, and the quantisation is imperceptible at macroscopic scales.