A nucleus at rest undergoes a decay emitting an alpha-particle of de-Broglie wavelength, lambda = 5.

Q: Why do both particles have the same momentum?

Since the parent nucleus was initially at rest, the initial momentum of the system is zero. By the law of conservation of linear momentum, the final total momentum must also be zero. Therefore, the two emitted particles must have equal and opposite momenta.

Q: Why do we use $p^2/2m$ instead of $1/2 mv^2$ for kinetic energy?

While both formulas are correct, we already calculated the momentum $p$ from the de-Broglie wavelength. Using $p^2/2m$ allows us to directly substitute the known momentum without having to calculate the velocities of the particles separately.

Q: Why did we convert the kinetic energy from Joules to MeV?

The masses of the nuclei are given in atomic mass units (amu), and the conversion factor provided is $1 \text{ amu} = 931.470 \text{ MeV}/c^2$. Converting the energy to MeV makes it straightforward to calculate the mass defect in amu.

Q: What is mass defect and why does it occur?

Mass defect is the difference in mass between the parent nucleus and the sum of the masses of the decay products. This 'missing' mass is converted into the kinetic energy of the products, as described by Einstein's equation $E = \Delta m c^2$.

Q: Can we ignore the kinetic energy of the parent nucleus?

Yes, the problem states that the parent nucleus is 'at rest', meaning its initial velocity and kinetic energy are exactly zero.

Question

A nucleus at rest undergoes a decay emitting an $\alpha$-particle of de-Broglie wavelength, $\lambda = 5.76 	imes 10^{-15} 	ext{ m}$. If the mass of the daughter nucleus is $223.610 	ext{ amu}$ and that of the $\alpha$-particle is $4.002 	ext{ amu}$. Determine the total kinetic energy in the final state. Hence obtain the mass of the parent nucleus in amu. ($1 	ext{ amu} = 931.470 	ext{ MeV}/c^2$)

Sigma Percentile · Accepted Answer

The phenomenon of radioactive decay is a beautiful demonstration of the fundamental laws of physics working in perfect harmony. In this problem, we are going to act as nuclear detectives, using the clues left behind by an emitted $\alpha$-particle to reconstruct the properties of the original parent nucleus.

### The Scene of the Decay

Imagine a heavy parent nucleus sitting perfectly at rest. Its initial momentum is zero. Suddenly, it undergoes alpha decay, splitting into a daughter nucleus and an $\alpha$-particle.

Because there are no external forces acting on the system, the **law of conservation of linear momentum** dictates that the total momentum must remain zero. This means the daughter nucleus and the $\alpha$-particle must fly apart in opposite directions with the exact same magnitude of momentum:
$$|\vec{p}_D| = |\vec{p}_\alpha| = p$$

### Decoding the Momentum

We are given a crucial piece of evidence: the de-Broglie wavelength of the $\alpha$-particle, $\lambda = 5.76 	imes 10^{-15} 	ext{ m}$. According to wave-particle duality, we can find the momentum $p$ using Planck's constant $h$:
$$p = \frac{h}{\lambda}$$

Substituting the known values:
$$p = \frac{6.63 	imes 10^{-34}}{5.76 	imes 10^{-15}} = 1.151 	imes 10^{-19} 	ext{ kg m/s}$$

Since momentum is conserved, this is also the momentum of the recoiling daughter nucleus!

### Calculating the Total Kinetic Energy

Now that we have the momentum, we can find the total kinetic energy $K$ of the system. The total kinetic energy is simply the sum of the kinetic energies of the two fragments. Using the relation $K = \frac{p^2}{2m}$, we get:
$$K = K_\alpha + K_D = \frac{p^2}{2m} + \frac{p^2}{2M}$$
$$K = \frac{p^2}{2} \left( \frac{1}{m} + \frac{1}{M} ight)$$

Let's plug in the masses (converting amu to kg using $1 	ext{ amu} = 1.67 	imes 10^{-27} 	ext{ kg}$) and the momentum we just found:
$$K = \frac{(1.151 	imes 10^{-19})^2}{2} \left( \frac{4.002 + 223.610}{4.002 	imes 223.610 	imes 1.67 	imes 10^{-27}} ight)$$
$$K \approx 10^{-12} 	ext{ J}$$

To make this value useful for nuclear calculations, we convert it to Mega electron-volts (MeV) by dividing by $1.6 	imes 10^{-13} 	ext{ J/MeV}$:
$$K = \frac{10^{-12}}{1.6 	imes 10^{-13}} = \mathbf{6.25 	ext{ MeV}}$$

### The Missing Mass

Where did this $6.25 	ext{ MeV}$ of kinetic energy come from? It didn't just appear out of nowhere. It came from the **mass defect** ($\Delta m$) of the reaction! During the decay, a tiny amount of the parent nucleus's mass was converted into pure energy, governed by Einstein's legendary equation, $\Delta E = \Delta m c^2$.

We can find this mass defect by dividing the total kinetic energy by the given conversion factor ($1 	ext{ amu} = 931.470 	ext{ MeV}/c^2$):
$$\Delta m = \frac{6.25 	ext{ MeV}}{931.470 	ext{ MeV/amu}} = 0.0067 	ext{ amu}$$

### Reconstructing the Parent Nucleus

Finally, we can determine the mass of the original parent nucleus. It must be equal to the sum of the masses of the decay products plus the mass that was converted into energy:
$$M_{	ext{parent}} = M + m + \Delta m$$
$$M_{	ext{parent}} = 223.610 + 4.002 + 0.0067$$
$$M_{	ext{parent}} = \mathbf{227.62 	ext{ amu}}$$

And there we have it! By simply observing the wavelength of the emitted $\alpha$-particle, we successfully deduced the total energy released and the exact mass of the original parent nucleus.

Visualized Solution

$p_{initial} = 0$

$p_{D} + p_{α} = 0$

$λ = \frac{h}{p}$

$p = 1.151 \times 1 0^{- 19} kg m/s$

$K = \frac{p ^{2}}{2 m} + \frac{p ^{2}}{2 M}$

$K = 6.25 MeV$

$Δ m = \frac{K}{c ^{2}}$

$Δ m = 0.0067 amu$

$M_{parent} = 227.62 amu$

The Sigma Insight: Radioactivity

The Scene of the Decay

Decoding the Momentum

Calculating the Total Kinetic Energy

The Missing Mass

Reconstructing the Parent Nucleus

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } pinitial​=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p​D​+p​α​=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=ph​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=1.151×10−19 kg m/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K=2mp2​+2Mp2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K=6.25 MeV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δm=c2K​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δm=0.0067 amu

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mparent​=227.62 amu