Suppose a 226_88 Ra nucleus at rest and in ground state undergoes alpha-decay to a 222_86 Rn nucleus

Q: Why do we use the ground state mass of Radon to find the Q-value?

The Q-value represents the total energy released in the entire process from the initial state to the final stable state. By using the ground state mass of Radon, we calculate the absolute maximum energy available, which includes both the kinetic energies and the gamma photon energy.

Q: Why doesn't the alpha particle get all the kinetic energy?

Because of the conservation of momentum. Since the initial Radium nucleus was at rest, the alpha particle and the Radon nucleus must move in opposite directions. The heavy Radon nucleus recoils, taking a small but significant portion of the kinetic energy.

Q: What is the formula $K_\alpha = \frac{A-4}{A} Q_{net}$ derived from?

It comes from combining the conservation of momentum ($p_\alpha = p_{\text{Rn}}$) and the kinetic energy formula ($K = \frac{p^2}{2m}$). Since they have the same momentum, their kinetic energies are inversely proportional to their masses.

Q: Why is the recoil energy of Radon so small?

Kinetic energy for a given momentum is inversely proportional to mass ($K = \frac{p^2}{2m}$). Since Radon is much more massive than the alpha particle (222 vs 4), it receives a proportionally much smaller share of the kinetic energy.

Q: Can we ignore the recoil of the nucleus when the gamma photon is emitted?

Yes, for all practical purposes in this context. The momentum of a gamma photon is extremely small compared to the massive Radon nucleus, so the recoil energy imparted by the gamma emission is negligible compared to the MeV scale energies we are dealing with.

Question

Suppose a $^{226}_{88}	ext{Ra}$ nucleus at rest and in ground state undergoes $\alpha$-decay to a $^{222}_{86}	ext{Rn}$ nucleus in its excited state. The kinetic energy of the emitted $\alpha$ particle is found to be $4.44	ext{ MeV}$. $^{222}_{86}	ext{Rn}$ nucleus then goes to its ground state by $\gamma$-decay. The energy of the emitted $\gamma$-photon is _______ $	ext{keV}$,
[Given: atomic mass of $^{226}_{88}	ext{Ra} = 226.005	ext{ u}$, atomic mass of $^{222}_{86}	ext{Rn} = 222.000	ext{ u}$, atomic mass of $\alpha$ particle = $4.000	ext{ u}$, $1	ext{ u} = 931	ext{ MeV/c}^2$, $c$ is speed of the light]

Sigma Percentile · Accepted Answer

### The Nuclear Detective Story

Imagine a Radium nucleus ($^{226}_{88}	ext{Ra}$) sitting perfectly at rest in a laboratory. It is highly unstable, a ticking quantum time bomb. Spontaneously, it spits out an alpha particle ($\alpha$) and transforms into a Radon nucleus ($^{222}_{86}	ext{Rn}$). But the story doesn't end there! The newly formed Radon nucleus is born in an **excited state**. To finally reach its stable ground state, it must release its excess energy as a gamma photon ($\gamma$).

Our mission today is to act as nuclear detectives and find the exact energy of this emitted photon, $E_\gamma$.

### The Energy Budget: Calculating the Q-Value

First, we need to calculate the $Q$-value of this entire process. The $Q$-value is simply the total energy released, which comes directly from the mass defect ($\Delta m$). According to Einstein's famous equation, $E = mc^2$, the missing mass is converted into pure energy.

We calculate the mass defect by subtracting the final ground-state masses from the initial mass:

$$Q = (m_{	ext{Ra}} - m_{	ext{Rn}} - m_\alpha) 	imes 931	ext{ MeV}$$

Let's substitute the given values. The mass of Radium is $226.005	ext{ u}$. From this, we subtract the Radon mass of $222.000	ext{ u}$, and the alpha mass of $4.000	ext{ u}$. Notice how clean these numbers are?

$$Q = (226.005 - 222.000 - 4.000) 	imes 931$$

Subtracting these gives a tiny mass defect of $0.005	ext{ u}$. Multiplying this by $931$ gives a total $Q$-value of **$4.655	ext{ MeV}$**. This is our total energy budget available for the entire decay process.

### Sharing the Spoils: Momentum Conservation

Now, how is this total energy distributed? It is shared among the kinetic energy of the alpha particle ($K_\alpha$), the recoiling Radon ($K_{	ext{Rn}}$), and the gamma photon ($E_\gamma$).

$$Q = K_{	ext{Rn}} + K_\alpha + E_\gamma$$

Because the initial Radium nucleus was at rest, the alpha particle and the Radon nucleus must fly apart in opposite directions to conserve momentum. Using momentum conservation, the alpha particle takes the lion's share of the available kinetic energy. The formula for the alpha particle's kinetic energy is the mass of the daughter nucleus divided by the parent nucleus, multiplied by the total available kinetic energy ($Q - E_\gamma$):

$$K_\alpha = \frac{A_{	ext{Rn}}}{A_{	ext{Ra}}} (Q - E_\gamma)$$

### The Final Calculation

We are given that the alpha particle has an energy of $4.44	ext{ MeV}$. The mass number of Radon is $222$, and Radium is $226$. Let's plug these into our equation:

$$4.44 = \frac{222}{226} (4.655 - E_\gamma)$$

To isolate $E_\gamma$, we multiply $4.44$ by the fraction $\frac{226}{222}$. This calculates to exactly $4.520	ext{ MeV}$. This is the total kinetic energy shared between the alpha particle and the recoiling Radon.

$$4.655 - E_\gamma = 4.520$$

Finally, we subtract this kinetic energy from our total $Q$-value.

$$E_\gamma = 4.655 - 4.520 = 0.135	ext{ MeV}$$

To convert this to kilo electron-volts, we multiply by $1000$, giving us our final answer of **$135	ext{ keV}$**. A beautiful application of mass-energy equivalence and momentum conservation!

Enter Numerical Value:

Visualized Solution

$_{88}^{226} Ra \to_{86}^{222} Rn^{*} + α$

$Q = (Δ m) c^{2}$

$Q = (226.005 - 222.000 - 4.000) \times 931$

$Q = 4.655 MeV$

$K_{α} = \frac{A _{Rn}}{A _{Ra}} (Q - E_{γ})$

$4.44 = \frac{222}{226} (4.655 - E_{γ})$

$4.655 - E_{γ} = 4.520$

$E_{γ} = 135 keV$

The Sigma Insight: Radioactivity

The Nuclear Detective Story

The Energy Budget

Sharing the Spoils

The Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 88226​Ra→86222​Rn∗+α

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q=(Δm)c2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q=(226.005−222.000−4.000)×931

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q=4.655 MeV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kα​=ARa​ARn​​(Q−Eγ​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4.44=226222​(4.655−Eγ​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4.655−Eγ​=4.520

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Eγ​=135 keV