Question bank — Nuclear reactions — Q-value, cross-section

False. $Q>0$ only says energy would be released; a charged projectile may still need to climb (or tunnel through) the Coulomb barrier to get close enough, so probability (the $\sigma$ question) is separate from the energy (the $Q$ question).

True (in principle for neutral projectiles). With $Q>0$ the products can already move as required, so $E_{th}=-Q(\frac{m_a+m_X}{m_X})<0$ is meaningless — no minimum kinetic energy is demanded; a charged projectile still faces a Coulomb barrier, which is a separate effect.

False. $Q$ is a difference of rest masses times $c^2$ — a frame-independent quantity. Only how the kinetic energy is shared changes between frames, not $Q$ itself.

False. Heavier products mean mass was gained, so $Q=\Delta (\text{rest mass})c^2<0$: energy must be supplied. Only mass lost becomes released kinetic energy.

True. $\sigma$ is a quantum probability measure; at a resonance it can exceed the physical area by thousands of times (and far below threshold it can nearly vanish), because it counts how likely the reaction is, not how big the nucleus is.

False. $\sigma$ is a property of the nucleus–projectile pair at a given energy; it does not know about beam intensity. Doubling $\varphi$ doubles the rate $R=\sigma \varphi$, leaving $\sigma$ unchanged.

False. An exponential never reaches zero; it only gets arbitrarily small. There is always a residual (tiny) transmitted intensity, which is why shielding is specified by an attenuation factor, not a "stopping distance."

False. Momentum conservation forces the products to keep drifting forward, locking up some energy as centre-of-mass motion; you need $E_{th}=|Q|\frac{m_a+m_X}{m_X}>|Q|$ (Picture 2).

False. Quantum tunnelling lets the cross-section stay small but nonzero below the barrier (Picture 3); this sub-barrier reaction rate is what powers fusion in stars.

False (if used consistently). For $(a,b)$ reactions that conserve total $Z$, the electron masses on both sides cancel, so consistent atomic masses give the correct $Q$; the error only appears when you mix nuclear and atomic masses.

False. Denser material means larger $n$, hence larger $\Sigma =n\sigma$, hence shorter mean free path — particles find targets sooner.

The flaw is ignoring momentum: the products cannot be created at rest, so part of the input energy is unavailable, giving $E_{th}=|Q|\frac{m_a+m_X}{m_X}$ (the recoil tax of Picture 2).

The number is an effective area from quantum resonance, not the geometric size; a real nucleus is roughly $10^{-24}{\text{cm}}^2$, so $755$ b vastly overstates its physical extent.

The binding-energy formula is reversed: $Q=(B_Y+B_b)-(B_X+B_a)$, and the reaction is exoergic when the products are more tightly bound (higher on the binding-energy curve of Picture 1).

Dimensionally and physically wrong: rate = area × incoming flux, so $R=\sigma \varphi$ (a product), giving reactions per target per unit time.

$\Sigma =n\sigma$ multiplies (nuclei per volume) by (area), giving inverse length (${\text{cm}}^{-1}$); barns are the unit of the microscopic $\sigma$ alone.

Reversed: for a heavy target $\frac{m_X}{m_a+m_X}\to 1$, so almost all lab energy is available in the CM frame — a heavy target barely recoils and wastes the least, so $E_{th}\to |Q|$.

Higher binding energy per nucleon means the nucleus is held together more strongly, i.e. more stable; that is precisely why moving toward the peak of the curve releases energy.

Both processes move nucleons toward the peak of the binding-energy-per-nucleon curve (Picture 1), raising total binding energy, so $Q=(B_{\text{products}})-(B_{\text{reactants}})>0$.

Nuclear rearrangements change rest mass measurably (fractions of a u), and that lost mass reappears as kinetic energy via $E=m{c}^2$ — far larger than chemical energies, which involve negligible mass change.

Only energy in the CM frame is free to be spent on rearranging nuclei; the CM's overall drift motion (fixed by momentum conservation) is untouchable, so the reaction "turns on" when $E_{cm}=|Q|$ (Picture 2).

The centre of mass is dragged forward by the projectile at $v_{cm}=\frac{m_a}{m_a+m_X}{v}_a$; the energy tied up in that forced drift is the fraction $\frac{m_a}{m_a+m_X}$, so the remainder $\frac{m_X}{m_a+m_X}$ is what stays available — the target's "share" of the mass sets how little is wasted.

A large $\sigma$ raises $\Sigma =n\sigma$, and half-thickness $=\frac{\ln 2}{\Sigma }$ shrinks as $\Sigma$ grows — so strongly-absorbing nuclei (like ${}^{10}\text{B}$ for neutrons) stop a beam in a very thin layer.

Each nucleus independently presents area $\sigma$ to flux $\varphi$, so one target reacts at rate $\sigma \varphi$; with $N$ identical independent targets the rates simply add, giving $N\sigma \varphi$.

$\sigma$ only tells you how big one target looks; the rate also needs how many projectiles actually arrive per second per area, and that arrival count is (how densely packed the projectiles are, $n$) times (how fast they sweep past, $v$) — so both a denser and a faster beam deliver more hits even though $\sigma$ is unchanged. See Neutron flux & reactor physics.

Neutrons feel no Coulomb repulsion, so even slow (thermal) neutrons reach the nucleus, while charged projectiles must first climb — or quantum-tunnel through — the Coulomb barrier (Picture 3), an energy hurdle separate from $Q$ or $E_{th}$.

Each thin slice removes a fixed fraction $n\sigma dx$ of whatever intensity remains, and "constant fractional loss per step" integrates to an exponential — the same logic as radioactive decay.

The formula gives a negative number, which we read as "no threshold": an exoergic reaction has no minimum kinetic energy requirement (a neutral projectile can react even at rest-like energies).

The factor $\frac{m_a+m_X}{m_X}\to 1$, so $E_{th}\to |Q|$: with a heavy, essentially immovable target almost no energy is lost to recoil, and you need barely more than $|Q|$.

It does not snap to zero — quantum tunnelling keeps it small but finite (Picture 3), which is why sub-barrier fusion in stars proceeds at all.

That channel's $\sigma$ is zero (the reaction is energetically forbidden), so no matter how intense the beam, the $(a,b)$ product never forms until $E_{th}$ is reached.

$\sigma$ spikes (possibly thousands of barns), so $\Sigma =n\sigma$ is large and the mean free path $1/\Sigma$ becomes very short — the beam is absorbed almost immediately.

$Q=0$: no mass converts to kinetic energy, so the reaction neither releases nor absorbs energy and behaves like an elastic rearrangement.

$\Sigma =n\sigma \to 0$ so $I\to I_0$: with nothing to hit, the beam passes through undiminished — the formula correctly reduces to "no attenuation."