The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction 16_7N + 4_2He -

Q: Why can't the alpha particle just have 1.86 MeV of kinetic energy?

If the alpha particle had exactly 1.86 MeV, it would use all of it to create the extra mass of the products. The products would then be at rest, meaning the final momentum is zero. But the alpha particle had initial momentum! This violates the conservation of momentum.

Q: Why do we assume the products move together at the threshold?

Any relative motion between the products (flying apart) would require additional kinetic energy. To find the absolute *minimum* energy required to make the reaction happen, we assume no energy is 'wasted' on relative motion, meaning they move together at the center of mass velocity.

Q: Can we use the approximate mass numbers (4 and 16) for momentum conservation?

Yes! While calculating the mass defect requires extreme precision (up to 3 decimal places) because the difference is tiny, the momentum and kinetic energy fractions depend on the mass ratios. Using 4 and 16 instead of 4.003 and 16.006 introduces a negligible error in the velocity ratio, making the algebra much cleaner.

Q: What happens if the alpha particle has more than 2.325 MeV?

The reaction will still occur, but now there is 'excess' energy. This extra energy will be converted into the relative kinetic energy of the Hydrogen and Oxygen nuclei, meaning they will fly apart from each other in the center of mass frame.

Q: Is the threshold energy formula applicable to all endothermic reactions?

Yes, the formula $K_{\text{th}} = |Q| (1 + m_{\text{projectile}} / m_{\text{target}})$ is universally applicable for any two-body collision where the target is initially at rest, provided the speeds are non-relativistic.

Question

The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction $^{16}_{7}	ext{N} + ^{4}_{2}	ext{He} ightarrow ^{1}_{1}	ext{H} + ^{19}_{8}	ext{O}$ in a laboratory frame is $n$ (in $	ext{MeV}$). Assume that $^{16}_{7}	ext{N}$ is at rest in the laboratory frame. The masses of $^{16}_{7}	ext{N}$, $^{4}_{2}	ext{He}$, $^{1}_{1}	ext{H}$ and $^{19}_{8}	ext{O}$ can be taken to be $16.006	ext{ u}$, $4.003	ext{ u}$, $1.008	ext{ u}$ and $19.003	ext{ u}$, respectively, where $1	ext{ u} = 930	ext{ MeVc}^{-2}$. The value of $n$ is_________.

Sigma Percentile · Accepted Answer

### The Anatomy of a Nuclear Reaction

Imagine you are observing a microscopic battlefield. An alpha particle ($^{4}_{2}	ext{He}$) is hurled at a stationary Nitrogen nucleus ($^{16}_{7}	ext{N}$). It's not just a simple bounce; it's an alchemical transformation where Nitrogen and Helium fuse and shatter into Hydrogen ($^{1}_{1}	ext{H}$) and Oxygen ($^{19}_{8}	ext{O}$). But this transformation isn't free. It demands an energy toll.

### Calculating the Energy Toll (The Q-Value)

In any nuclear reaction, we must meticulously account for mass. Let's calculate the mass defect ($\Delta m$). We subtract the total mass of the reactants from the total mass of the products:

$$\Delta m = (m_{	ext{H}} + m_{	ext{O}}) - (m_{	ext{N}} + m_{\alpha})$$

Substituting the given values:

$$\Delta m = (1.008 + 19.003) - (16.006 + 4.003) = 20.011 - 20.009 = 0.002	ext{ u}$$

The products weigh slightly more than the reactants! Where does this extra mass come from? Einstein's $E = mc^2$ tells us it comes from the kinetic energy of the incoming alpha particle. Multiplying the mass defect by $930	ext{ MeV/u}$, we find the energy required, known as the Q-value:

$$|Q| = 0.002 	imes 930 = 1.86	ext{ MeV}$$

### The Hidden Cost: Conservation of Momentum

Here is where most students fall into a trap. They assume the alpha particle only needs $1.86	ext{ MeV}$ of kinetic energy. But wait! If the alpha particle gives up all its kinetic energy to create mass, the resulting Hydrogen and Oxygen would be perfectly still.

This violates the sacred law of conservation of momentum! The alpha particle came in with initial momentum, so the products must carry that exact same momentum forward. Therefore, the alpha particle must provide the Q-value **PLUS** the kinetic energy of the products moving together.

### The Threshold Condition

To find the *minimum* kinetic energy, we look at the threshold condition. At this absolute minimum, the products don't have any spare energy to fly apart from each other. They move together as a single clump, coasting along at the velocity of the center of mass ($v_c$).

Let's apply conservation of linear momentum. The initial momentum is the mass of the alpha particle times its velocity $v_0$. The final momentum is the total mass times $v_c$. Using the approximate mass numbers ($4$ and $16$) for simplicity:

$$m_{\alpha} v_0 = (m_{\alpha} + m_{	ext{N}}) v_c$$

$$4 v_0 = (4 + 16) v_c \implies 4 v_0 = 20 v_c \implies v_c = \frac{v_0}{5}$$

### The Mathematical Execution

Now, let's write the energy conservation equation. The initial kinetic energy ($K$) of the alpha particle goes into the Q-value and the final kinetic energy of the combined products:

$$K = |Q| + \frac{1}{2} (m_{\alpha} + m_{	ext{N}}) v_c^2$$

$$K = 1.86 + \frac{1}{2} (20) \left(\frac{v_0}{5}ight)^2$$

Simplifying the final kinetic energy term, it becomes $\frac{2}{5} v_0^2$. We know the initial kinetic energy $K = \frac{1}{2} (4) v_0^2 = 2 v_0^2$, which means $v_0^2 = \frac{K}{2}$. Substituting this back:

$$K = 1.86 + \frac{2}{5} \left(\frac{K}{2}ight) = 1.86 + \frac{K}{5}$$

Bringing the $K$ terms to one side:

$$K - \frac{K}{5} = 1.86 \implies \frac{4K}{5} = 1.86$$

$$K = \frac{1.86 	imes 5}{4} = 2.325	ext{ MeV}$$

### The Master Formula

You can also use a direct, elegant formula for threshold kinetic energy. It bypasses the algebra entirely:

$$K_{	ext{th}} = |Q| \left( 1 + \frac{m_{	ext{projectile}}}{m_{	ext{target}}} ight)$$

Substituting our values:

$$K_{	ext{th}} = 1.86 \left( 1 + \frac{4}{16} ight) = 1.86 	imes \frac{5}{4} = 2.325	ext{ MeV}$$

Always keep this formula handy for quick execution in exams!

Enter Numerical Value:

Visualized Solution

$The Nuclear Reaction$

$Δ m (Mass Defect)$

$∣ Q ∣ (Energy Required)$

$Threshold Kinetic Energy Concept$

$Conservation of Momentum$

$Conservation of Energy$

$Solving for Minimum Kinetic Energy$

$Final Calculation$

$The Direct Formula (Alternative)$

The Sigma Insight: Nucleus and Nuclear Reaction

The Anatomy of a Nuclear Reaction

Calculating the Energy Toll (The Q-Value)

The Hidden Cost

The Threshold Condition

The Mathematical Execution

The Master Formula

Similar Questions

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Comprehension Passage

The correct statement is

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Assume that a neutron breaks into a proton and an electron. The energy released during this process is (mass of neutron $= 1.6725 \times 1 0^{- 27}$ kg, mass of proton $= 1.6725 \times 1 0^{- 27}$ kg, mass of electron $= 9 \times 1 0^{- 31}$ kg)

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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; } The Nuclear Reaction

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Threshold Kinetic Energy Concept

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Momentum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Minimum Kinetic Energy

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Direct Formula (Alternative)