Alpha Beta and Gamma Decay Questions and Answers

Answer: B

The match is 1-iv, 2-i, 3-ii, 4-iii.

The order of forbiddenness is set by the spin change \(\Delta I\) and whether the parity changes. Allowed decays have \(\Delta I = 0, 1\) and no parity change. A forbidden decay of order \(n\) needs \(n\) units of orbital angular momentum, with parity change for odd \(n\); a unique transition has \(\Delta I = n + 1\).

• 1: \(\frac{1}{2}^{+} \to \frac{1}{2}^{+}\), \(\Delta I = 0\), no parity change: allowed.
• 2: \(\frac{1}{2}^{-} \to \frac{5}{2}^{+}\), \(\Delta I = 2\), parity changes: first forbidden (unique).
• 3: \(3^{+} \to 0^{+}\), \(\Delta I = 3\), no parity change: second forbidden (unique).
• 4: \(4^{-} \to 0^{+}\), \(\Delta I = 4\), parity changes: third forbidden (unique).

The other options misplace the allowed decay or reverse the order of the forbidden ones.

Answer: C

The correct statement is \(2^{+} \rightarrow 0^{+}\) is an electric quadrupole transition.

A photon of multipolarity \(L\) must satisfy \(|J_{i} - J_{f}| \le L \le J_{i} + J_{f}\) (and \(L \ge 1\)). An electric \(EL\) transition changes parity by \((-1)^{L}\) and a magnetic \(ML\) transition by \((-1)^{L+1}\).

For \(2^{+} \rightarrow 0^{+}\), only \(L = 2\) is allowed, and there is no parity change, so it must be E2, electric quadrupole. M2 would need a parity change.

For \(4^{+} \rightarrow 2^{+}\), \(L\) runs from 2 to 6 with no parity change, so the allowed multipoles are E2, M3, E4, M5, E6, and E2 dominates. An electric octupole (E3) or a magnetic quadrupole (M2) would both change parity, so the first two statements are wrong, as is the last.

Answer: C

The correct distribution is the continuous curve that starts above zero, peaks at low energy and falls to zero (C).

In beta decay the available energy is shared among three bodies: the electron, the antineutrino and the recoiling nucleus. So the electron energy is not fixed but spread continuously from 0 up to the endpoint energy, which rules out the single line of (A), the shape expected for a two-body decay.

For electron (β^-) emission the Coulomb attraction of the daughter nucleus slows the electrons, piling them up at low energy; the number at very low energy stays finite rather than vanishing, and the spectrum then falls off towards the endpoint. Curve (C) has exactly this shape.

A symmetric bell curve that vanishes at zero energy (B) ignores that Coulomb effect, and a curve peaking near the maximum energy (D) would mean the neutrino usually carries almost nothing, which is not what is observed.

Answer: B

The correct option is \(X_{1} \to Y_{1}\) is a Fermi transition and \(X_{2} \to Y_{2}\) is a Gamow-Teller transition.

In allowed beta decay there is no change of parity, and the selection rules depend on how the electron and neutrino spins combine. In a Fermi transition their spins are antiparallel (total 0), so \(\Delta J = 0\). In a Gamow-Teller transition they are parallel (total 1), so \(\Delta J = 0, \pm 1\), except that \(0 \to 0\) is forbidden.

\(0^{+} \to 0^{+}\) (\(X_{1} \to Y_{1}\)) cannot be Gamow-Teller, so it is a pure Fermi transition. \(0^{+} \to 1^{+}\) (\(X_{2} \to Y_{2}\)) has \(\Delta J = 1\), which a Fermi transition cannot provide, so it is pure Gamow-Teller.

That rules out every other combination. (This is a multiple-select question; the key accepts this option alone.)

Answer: A

The barrier height is about 26 MeV.

The barrier is the Coulomb energy of the alpha particle (charge 2e) touching the daughter nucleus (charge 82e). With \( e^{2}/4\pi\epsilon_{0} = 1.44 \) MeV fm and \( r_{0} \approx 1.2 \) fm:

\[ \begin{aligned} R &= 1.2\,(206^{1/3} + 4^{1/3}) \approx 9.0\ \text{fm} \\ V &= \frac{1.44 \times 2 \times 82}{9.0} \approx 26\ \text{MeV} \end{aligned} \]

The alpha particles of Po-210 have only about 5·3 MeV, far below this, which is why alpha decay proceeds by quantum tunnelling. 40, 42 and 80 MeV would need a much smaller contact radius.

Answer: B

The decay is a second forbidden Gamow-Teller transition.

The spin changes by ΔJ = 3 (from 3 to 0, the paper's O⁺ being 0⁺), and the parity does not change (+ to +). The classification is:

• Allowed: ΔJ = 0, ±1, no parity change (Fermi for ΔJ = 0 with no spin flip, Gamow-Teller for ΔJ = 0, ±1).
• First forbidden: parity changes, ΔJ up to 2.
• Second forbidden: no parity change, ΔJ = 2 or 3.

ΔJ = 3 with no parity change needs two units of orbital angular momentum plus the electron and neutrino spins aligned, which is a second forbidden (unique) Gamow-Teller decay. An allowed Fermi or Gamow-Teller decay cannot change J by 3, and a first forbidden decay requires a parity change.

Answer: B

The end point energy is nearly 0.84 MeV.

With atomic masses, the Q-value of positron decay must subtract two electron masses: one for the emitted positron and one for the extra atomic electron of the parent:

\[ \begin{aligned} \Delta M c^{2} &= (79.918528 - 79.916520) \times 931.5 \\ &= 0.002008 \times 931.5 = 1.870\ \text{MeV} \\ Q &= 1.870 - 1.022 = 0.848\ \text{MeV} \end{aligned} \]

The maximum (end point) energy of the positron is this Q-value, about 0.84 MeV. Forgetting the 1.022 MeV gives 1.87 MeV, close to option C, while 0.511 and 1.022 MeV are electron rest-mass energies.

Answer: D

The average beta energy is about 4.3 KeV.

Convert the heat to joules and divide by the number of decays:

\[ \begin{aligned} \bar{E} &= \frac{21 \times 4.186\ \text{J}}{1.27 \times 10^{17}} = 6.92 \times 10^{-16}\ \text{J} \\ &= \frac{6.92 \times 10^{-16}}{1.6 \times 10^{-19}}\ \text{eV} \approx 4.3\ \text{keV} \end{aligned} \]

As an average it lies well below the 18.6 keV end point of tritium decay, as it must, since the beta spectrum is continuous. The larger options would need more heat or fewer decays.

Answer: C

The decay is possible because the mass of \(^{5}\)Li is greater than the sum of the masses of \(^{4}\)He and P.

A spontaneous decay can occur only if it releases energy, that is if the Q-value is positive:

\[ \begin{aligned} Q &= [M(^{5}\text{Li}) - M(^{4}\text{He}) \\ &\quad - M_{p}]c^{2} \gt 0 \end{aligned} \]

Lithium-5 is indeed unbound: it falls apart into an alpha particle and a proton almost instantly, with a Q-value of about 2 MeV. If its mass were smaller than the products' mass, the decay would be energetically forbidden. Spin decides which angular momenta the products may carry away, not whether the decay is possible, so the options about spin are wrong.

Answer: B

The possible multipoles are M1, E2, M3, E4.

A gamma ray of multipole order L must satisfy \(|J_{i} - J_{f}| \le L \le J_{i} + J_{f}\) with \(L \ge 1\):

\[ \begin{aligned} L_{min} &= \left|\frac{5}{2} - \frac{3}{2}\right| = 1 \\ L_{max} &= \frac{5}{2} + \frac{3}{2} = 4 \end{aligned} \]

Both states have negative parity, so there is no parity change. Electric multipoles EL change parity by \((-1)^{L}\) and magnetic ones ML by \((-1)^{L + 1}\), so with no change of parity the allowed modes are the even-L electric and odd-L magnetic ones: M1, E2, M3, E4 (in practice M1 and E2 dominate). E1, M2, E3, M4 would require a parity change, and options C and D include orders above 4.