Atoms and Nuclei Questions and Answers

Answer: B

After 20 minutes, 128 nuclei are left.

20 minutes is \(\frac{20}{4} = 5\) half-lives, and each halves the number of undecayed nuclei:

\[ \begin{aligned} N &= 4096 \times \left(\tfrac{1}{2}\right)^{5} \\ &= \frac{4096}{32} = 128 \end{aligned} \]

The other options stop too early: 1024 is after 2 half-lives (8 minutes), 512 after 3 (12 minutes) and 256 after 4 (16 minutes).

Answer: C

The key gives \(\bar{p} \propto A^{-1/3}\).

The key follows the uncertainty-principle estimate. A nucleon is confined within the nucleus, whose radius is \(R = R_{0}A^{1/3}\), so its momentum spread is of order

\[ \begin{aligned} \bar{p} &\sim \frac{\hbar}{R} = \frac{\hbar}{R_{0}A^{1/3}} \\ &\propto A^{-1/3} \end{aligned} \]

This is the textbook estimate the question intends; a fuller Fermi-gas treatment, which accounts for all A nucleons filling momentum states, gives a typical momentum that is nearly independent of A. Among the options, \(A^{1/3}\) has the dependence inverted, and the forms in Z or \(AZ\) have no basis, since the size of the nucleus depends on the total number of nucleons.

Answer: A

The rate is \(\frac{\alpha E_{0}}{m\,s}(1 - e^{-\lambda t})\).

The number \(N\) of nuclei of \(X\) grows by production and falls by decay:

\[ \begin{aligned} \frac{dN}{dt} &= \alpha - \lambda N,\quad N(0) = 0 \\ N &= \frac{\alpha}{\lambda}\left(1 - e^{-\lambda t}\right) \end{aligned} \]

The activity is \(\lambda N = \alpha(1 - e^{-\lambda t})\) decays per second, each releasing \(E_{0}\). With no heat loss this power goes into the liquid, \(P = ms\,\frac{dT}{dt}\), so \(\frac{dT}{dt} = \frac{\alpha E_{0}}{ms}(1 - e^{-\lambda t})\).

The physics checks the shape: at \(t = 0\) there are no nuclei, so the heating rate starts at zero; later the activity saturates at the production rate \(\alpha\). The \((e^{\lambda t} - 1)\) form grows without limit, which would need activity to exceed production. The \(\lambda E_{0}\) form saturates at the wrong value, since activity levels off at \(\alpha\), not \(\lambda\). The \((\alpha - \lambda e^{-\lambda t})\) form is not zero at \(t = 0\).

Answer: A

The ratio is \(\frac{5}{4}\).

Mean life and half-life are proportional: \(\tau = \frac{1}{\lambda}\) and \(T_{1/2} = \frac{\ln 2}{\lambda}\), so \(\tau = \frac{T_{1/2}}{\ln 2}\).

\[ \frac{\tau_{A}}{\tau_{B}} = \frac{1500}{1200} = \frac{5}{4} \]

The factor \(\ln 2\) cancels in the ratio. The other fractions, all less than 1, would make A shorter-lived even though its half-life is longer.

Answer: C

The greatest wavelength is about 910 Å.

Ionising a ground-state hydrogen atom needs 13.6 eV. The longest wavelength that can supply it is

\[ \lambda = \frac{hc}{E} = \frac{12400}{13.6} \approx 912 \]

in Ångstrom units, using \(hc \approx 12400\) eV Å.

This is the series limit of the Lyman series. Longer wavelengths such as 1400 or 1820 Å carry too little energy; shorter ones such as 450 or 700 Å also ionise, but the question asks for the greatest.

Answer: D

The atom emits 6 spectral lines.

After absorbing 12.75 eV, the electron's energy is \(-13.6 + 12.75 = -0.85\) eV, which is \(-13.6/n^{2}\) with \(n = 4\). From n = 4 it can return to the ground state by any combination of downward jumps, giving

\[ \frac{n(n - 1)}{2} = \frac{4 \times 3}{2} = 6 \]

lines (4→3, 4→2, 4→1, 3→2, 3→1, 2→1). 3 lines would be for excitation to n = 3, and 4 counts only the jumps from n = 4.

Answer: D

The position is lowered by two places.

An α-particle is a helium nucleus with 2 protons and 2 neutrons. Emitting it lowers the atomic number by 2 and the mass number by 4, so the daughter element lies two places to the left (earlier) in the periodic table. For example, uranium-238 (Z = 92) decays to thorium-234 (Z = 90).

Beta emission moves an element one place up (Z increases by 1), and gamma emission leaves it unchanged; neither describes α-decay.

Answer: E

The process that releases neutrons from the nucleus is nuclear fission.

When a heavy nucleus such as uranium-235 absorbs a neutron and splits, the fragments emit free neutrons, on average between two and three per fission. Those neutrons carry on the chain reaction in a reactor or a bomb, so fission is the standard answer.

The other options do not release free neutrons. In β-decay a neutron turns into a proton and the nucleus emits an electron and an antineutrino; α-decay emits a helium nucleus, whose neutrons leave bound inside it; pair production turns a photon into an electron and a positron. Nuclear fusion is the one close competitor: some fusion reactions (deuterium with deuterium or tritium) do give off a neutron, but others, such as the proton-proton fusion that powers the Sun, do not, so fusion is not a process that releases neutrons as a rule, and fission is the better answer.

Answer: C

The volume of the second nucleus is 4V.

Nuclear radius grows with the cube root of the mass number, \(R = R_{0}A^{1/3}\), so nuclear volume is proportional to the mass number itself. This reflects the near-constant density of nuclear matter.

\[ \begin{aligned} V &\propto R^{3} \propto A \\ \frac{V_{2}}{V_{1}} &= \frac{8}{2} = 4 \end{aligned} \]

2V is the ratio of the radii (\(8^{1/3}/2^{1/3} = 2\)), the trap if the cube-root rule is applied to volume instead of radius. 3V, 6V and 9V do not follow from volume being proportional to mass number.

Answer: A

The number decayed is \(\frac{31}{32}N\).

After each half-life the remaining number halves, so after five half-lives

\[ \begin{aligned} N_{left} &= N\left(\tfrac{1}{2}\right)^{5} = \frac{N}{32} \\ N_{decayed} &= N - \frac{N}{32} = \frac{31}{32}N \end{aligned} \]

\(\frac{N}{32}\) is the number left, the main trap, and \(\frac{15}{16}N\) and \(\frac{7}{8}N\) are the decayed fractions after four and three half-lives.