10 questions · 1 mark each · Correct answer highlighted with explanation
10 questions · 3 marks each · Full model answers included
Observations: Most alpha particles passed straight through the gold foil with negligible deflection; a small fraction were deflected at large angles; a very small fraction bounced almost straight back.
Conclusions: Since most particles passed through undeflected, most of the atom is empty space. Since some particles were deflected at large angles (or bounced back), a small, dense, positively charged region — the nucleus — must exist, concentrating the atom's positive charge and most of its mass in a tiny central volume.
According to the Bohr model, electrons can only occupy specific, discrete (quantised) energy levels within an atom. When an electron transitions from a higher energy level to a lower one, it emits a photon whose energy exactly equals the energy difference between these two specific levels: $E_{photon} = E_i - E_f = hf$.
Since only specific, discrete energy differences are possible (corresponding to transitions between the atom's specific allowed energy levels), only specific, discrete photon frequencies/wavelengths are emitted — this is precisely why atomic emission spectra consist of distinct, sharp spectral lines, rather than a continuous range of wavelengths.
Mass defect ($\Delta m$) is the difference between the total mass of the individual, separated nucleons (protons and neutrons) and the actual mass of the combined nucleus — the nucleus's mass is always slightly less than the sum of its individual component masses.
Binding energy is the energy equivalent of this mass defect, related by Einstein's mass-energy equivalence: $E_b = \Delta mc^2$ — representing the energy released when the nucleus forms (or equivalently, the energy that must be supplied to completely separate the nucleus into individual nucleons).
The binding energy per nucleon curve rises for light nuclei, peaks around iron (mass number ≈ 56), and then gradually decreases for heavier nuclei — meaning medium-mass nuclei (like iron) are the most stable, most tightly bound per nucleon.
This explains both nuclear processes releasing energy: fusion of very light nuclei (like hydrogen) into medium-mass nuclei moves along the rising part of the curve, increasing binding energy per nucleon (releasing energy). Fission of very heavy nuclei (like uranium) into medium-mass fragments moves along the falling part of the curve (in reverse), also increasing binding energy per nucleon (releasing energy) — both processes release energy specifically because they move nuclei towards the more stable, more tightly bound medium-mass region of the curve.
Alpha decay: Emission of an alpha particle (helium nucleus, 2p+2n) — mass number decreases by 4, atomic number decreases by 2.
Beta decay: Emission of a beta particle (electron, from neutron-to-proton conversion within the nucleus) — mass number unchanged, atomic number increases by 1 (for standard beta-minus decay).
Gamma decay: Emission of a high-energy gamma photon, as an already-formed nucleus transitions from an excited (higher energy) state to a lower energy state — no change in mass number or atomic number, since only energy (not particles) is emitted.
The law of radioactive decay states that the rate of decay of radioactive nuclei is directly proportional to the number of undecayed nuclei present: $N = N_0e^{-\lambda t}$, where $\lambda$ is the decay constant.
At half-life $T_{1/2}$, $N = N_0/2$: $\dfrac{N_0}{2} = N_0e^{-\lambda T_{1/2}} \Rightarrow \dfrac{1}{2} = e^{-\lambda T_{1/2}}$
Taking natural log: $\ln(1/2) = -\lambda T_{1/2} \Rightarrow T_{1/2} = \dfrac{\ln2}{\lambda} = \dfrac{0.693}{\lambda}$
Chemical reactions involve rearrangement of electrons in atomic bonds — the energy changes involved are on the order of a few electron volts (eV) per reaction event.
Nuclear reactions (fission or fusion) involve rearrangement of nucleons within the nucleus, governed by the much stronger nuclear force — the energy changes involved are on the order of millions of electron volts (MeV) per reaction event, roughly a million times greater than typical chemical reaction energies. This vast difference in energy scale (nuclear versus chemical/electronic binding forces) is precisely why nuclear reactions release such enormously greater energy per unit mass of fuel consumed, compared to chemical reactions.
A chain reaction occurs when the fission of one nucleus (like uranium-235) releases additional neutrons, which go on to trigger fission in further nearby nuclei, releasing yet more neutrons — this self-sustaining, potentially rapidly escalating process is a chain reaction.
Critical mass is the minimum mass of fissile material required for a self-sustaining chain reaction to occur — below this mass, too many neutrons escape without causing further fission (due to insufficient surrounding material), preventing the reaction from sustaining itself.
The Bohr model, while successfully explaining the hydrogen atom's spectral lines, has significant limitations — most notably, it fails to accurately explain the spectra of atoms with more than one electron (multi-electron atoms), since it does not properly account for the additional complexity of electron-electron interactions.
The Bohr model also cannot explain finer details even within hydrogen's spectrum (like fine structure splitting, or the effect of external magnetic/electric fields on spectral lines), and its somewhat ad-hoc combination of classical orbital mechanics with an arbitrarily imposed quantisation rule lacks a deeper theoretical justification — limitations that were only properly resolved by the subsequent, more complete development of full quantum mechanics.
10 questions · 5 marks each · Detailed step-by-step solutions
$E_2 = \dfrac{-13.6}{2^2} = \dfrac{-13.6}{4}$
$E_2 = \mathbf{-3.4\text{ eV}}$
$E_3 = -13.6/9 \approx -1.51\text{ eV}$; $E_2 = -13.6/4 = -3.4\text{ eV}$
Energy released: $\Delta E = E_3 - E_2 = -1.51-(-3.4) = 1.89\text{ eV}$
Converting to joules: $\Delta E = 1.89\times1.6\times10^{-19} = 3.024\times10^{-19}\text{ J}$
Wavelength: $\lambda = \dfrac{hc}{\Delta E} = \dfrac{6.63\times10^{-34}\times3\times10^8}{3.024\times10^{-19}} \approx \dfrac{1.989\times10^{-25}}{3.024\times10^{-19}}$
$\lambda \approx \mathbf{6.58\times10^{-7}\text{ m} = 658\text{ nm}}$ (this is the well-known H-alpha line in the Balmer series)
Number of half-lives elapsed: $n = 30/10 = 3$
Fraction remaining: $\left(\dfrac{1}{2}\right)^n = \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$
$\dfrac{1}{8}$ (12.5%) of the original sample remains after 30 days.
$\lambda = \dfrac{0.693}{T_{1/2}} = \dfrac{0.693}{5}$
$\lambda = \mathbf{0.1386\text{ per year} \approx 0.139\text{ yr}^{-1}}$
$E = \Delta m\times931.5\text{ MeV}$
$E = 0.02\times931.5$
$E = \mathbf{18.63\text{ MeV}}$
Number of half-lives: $n = 12/4 = 3$
Remaining nuclei: $N = N_0\left(\dfrac{1}{2}\right)^n = 8\times10^{10}\times\left(\dfrac{1}{2}\right)^3 = 8\times10^{10}\times\dfrac{1}{8}$
$N = \mathbf{1\times10^{10}\text{ nuclei}}$
$r_1 = 0.529\times1^2 = 0.529\text{ Å}$
$r_1 = \mathbf{0.529\times10^{-10}\text{ m} = 5.29\times10^{-11}\text{ m}}$
This is the well-known Bohr radius, the characteristic size of the hydrogen atom in its ground state.
Binding energy: $E_b = \Delta m\times931.5 = 0.3\times931.5 = 279.45\text{ MeV}$
Binding energy per nucleon: $\dfrac{E_b}{A} = \dfrac{279.45}{24} \approx \mathbf{11.64\text{ MeV/nucleon}}$
Activity follows the same decay law as the number of nuclei: $A = A_0\left(\dfrac{1}{2}\right)^n$
Number of half-lives: $n = 18/6 = 3$
$A = 8000\times\left(\dfrac{1}{2}\right)^3 = 8000\times\dfrac{1}{8}$
$A = \mathbf{1000\text{ disintegrations/second}}$
Total initial mass: $4\times1.008 = 4.032\text{ u}$
Final mass: $4.002\text{ u}$
Mass defect: $\Delta m = 4.032 - 4.002 = 0.03\text{ u}$
Energy released: $E = \Delta m\times931.5 = 0.03\times931.5$
$E \approx \mathbf{27.9\text{ MeV}}$
This is essentially the process powering the Sun and most stars — the fusion of hydrogen into helium, releasing substantial energy per reaction.