Alpha Decay Q-Value

Equivalent forms

Q_\alpha = T_\alpha\left(1 + \frac{m_\alpha}{m_D}\right)

A simple subtraction of three masses unlocks the energy that warms Earth's interior.

Symbol	Name	SI unit	Dimension	Range
$Q$	Q-valueoutput Energy released in the alpha decay	J	M·L²·T⁻²	0 – 2e-12
$Δm$	Mass defect Difference between parent rest mass and total product rest mass	kg	M	0 – 2e-29
$c$	Speed of light Speed of light in vacuum	m/s	L·T⁻¹	299792458 – 299792458

Unit systems

Symbol	SI	CGS	Natural
$Q$	J	erg	MeV
$Δm$	kg	g	MeV/c²
$c$	m/s	cm/s	1

Where it holds

Applicable to spontaneous alpha emission from heavy nuclei where Q > 0; recoil correction matters at percent level.

Dimensional analysis

Q \to [M\cdot L^{2}\cdot T^{-2}]

\Delta m\cdot c^{2} \to [M]\cdot [L\cdot T^{-1}]^{2} = [M\cdot L^{2}\cdot T^{-2}]

Both sides are Joules.

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Discovery

George Gamow · 1928

Gamow explained alpha decay via quantum tunneling, connecting the Q-value to half-life and barrier penetration.

Try this

How much energy does a uranium nucleus release when it spits out an alpha particle?

U-238 decays to Th-234 by emitting an alpha particle. Given the mass defect Δm = 7.57×10⁻³⁰ kg, compute the released energy in MeV.

Research status: stable

Real-world applications

Smoke detectors (Am-241 alpha source)
Radioisotope thermoelectric generators powering deep-space probes (Pu-238)
Earth's internal heat budget (U, Th alpha chains)
Targeted alpha therapy in oncology (Ac-225, Ra-223)

Common misconceptions

The alpha particle does not carry all the Q-value; the daughter nucleus recoils with a small fraction $(\sim m_\alpha /m_P)$ .
Q-value alone does not predict half-life — quantum tunneling probability through the Coulomb barrier sets the decay rate (Geiger-Nuttall).
Atomic vs nuclear masses must be handled consistently; electron binding corrections are usually negligible.

Experimental verification

Magnetic spectrometers measuring alpha kinetic energies (e.g., Rosenblum 1929) agree with Q-values computed from atomic mass tables to keV precision.

Derivation

Conservation of energy and momentum in the parent rest frame gives

p_\alpha = p_D

and

(m_P)c^{2} = \surd ((p_\alpha c)^{2} + (m_\alpha c^{2})^{2}) + \surd ((p_D c)^{2} + (m_D c^{2})^{2})

.

In the non-relativistic limit, T_total

= p^{2}/(2m_\alpha ) + p^{2}/(2m_D) = Q

, so

T_\alpha = Q \cdot m_D/(m_\alpha + m_D)

.

Limiting cases

\Delta m \to 0

⟶

Q \to 0

No spontaneous decay can occur.

\Delta m = m_\alpha

⟶

Q \approx 3727\,\mathrm{MeV}

Hypothetical complete conversion of the alpha rest mass.

\Delta m \approx 7.57e-30\,\mathrm{kg} (U-238)

⟶

Q \approx 4.27\,\mathrm{MeV}

Observed Q-value for

U-238 \to Th-234 + \alpha

.

What if…

What

if \Delta m

were negative?

Q would be negative and the decay would be energetically forbidden; the nucleus would be stable against alpha emission.

What if the daughter were infinitely heavy?

The alpha would carry essentially all of Q as kinetic energy with no measurable recoil.

What if Q doubled?

Half-life would plummet by many orders of magnitude (Geiger-Nuttall law: $\log T_{1}/_{2} \propto 1/\sqrt{Q})$ .

1

U-238 alpha decay Q-value

Given ·

Δm:: 7.57e-30
c:: 299792458

Find · Q

Steps

Step 1: $c^{2} = (299792458)^{2} = 8.988e16\,\mathrm{m}^{2}/s^{2}$ .
Step 2: $Q = 7.57e-30 \times 8.988e16 = 6.80e-13\,\mathrm{J}$ .
Step 3: Convert: $6.80e-13 / 1.602176634e-13 \approx 4.25\,\mathrm{MeV} (4.27$ with full precision masses).

Result ·

Q = \Delta m c^{2} = 7.57e-30 \times (299792458)^{2} \approx 6.80e-13\,\mathrm{J} \approx 4.27\,\mathrm{MeV}

2

Kinetic energy of the alpha (recoil-corrected)

Given ·

Q:: 6.8e-13
Δm:: 7.57e-30
c:: 299792458

Find · T_alpha

Steps

Step 1: $m_D \approx 234\,\mathrm{u}$ , $m_\alpha \approx 4\,\mathrm{u}$ .
Step 2: Fraction $= 234/238 \approx 0.9832$ .
Step 3: $T_\alpha \approx 0.9832 \times 4.27\,\mathrm{MeV} \approx 4.20\,\mathrm{MeV}$ .
Step 4: The remaining 0.07 MeV is the recoil kinetic energy of Th-234.

Result ·

T_\alpha = Q \cdot m_D/(m_\alpha + m_D) \approx 4.27 \times 234/238 \approx 4.20\,\mathrm{MeV}

Mass–Energy Equivalence→q value nuclear reactiongeiger nuttall law