Problem

Determine the mass of the lithium isotope ${}_{3}\mathrm{Li}^{6}$ when it is bombarded by deuterons, resulting in two alpha particles and the release of energy equal to $22.3 \, \mathrm{MeV}$.

The reaction is:

$${}_{1}^{2}\mathrm{H} + {}_{3}^{6}\mathrm{Li} \rightarrow 2 \, {}_{2}^{4}\mathrm{He} + Q$$

Data

• Mass of deuteron (${}_{1}^{2}\mathrm{H}$): $2.014 \, \mathrm{u}$
• Mass of alpha particle (${}_{2}^{4}\mathrm{He}$): $4.0026 \, \mathrm{u}$
• Energy released ($Q$): $22.3 \, \mathrm{MeV}$
• Conversion factor:
  • $1 \, \mathrm{u} = 931.5 \, \mathrm{MeV}$
  • $1 \, \mathrm{MeV} = \frac{1}{931.5} \, \mathrm{u}$

Prerequisite Concepts

1. Energy to Mass Conversion:
   The energy released ($Q$) can be expressed in atomic mass units ($\mathrm{u}$):

$$Q = \frac{22.3}{931.5} \, \mathrm{u}$$

2. Mass Conservation in Nuclear Reactions:
   The total mass of the reactants equals the total mass of the products plus the energy released.
   Rearranging the reaction equation:

$$m_{{}_{3}^{6}\mathrm{Li}} = 2 \, m_{{}_{2}^{4}\mathrm{He}} + Q - m_{{}_{1}^{2}\mathrm{H}}$$

Solution

Step 1: Convert Energy to Mass ($Q$)

$$Q = \frac{22.3}{931.5} \, \mathrm{u}$$ $$Q = 0.0239 \, \mathrm{u}$$

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Step 2: Substitute Values into the Mass Equation

Using the formula:

$$m_{{}_{3}^{6}\mathrm{Li}} = 2 \, m_{{}_{2}^{4}\mathrm{He}} + Q - m_{{}_{1}^{2}\mathrm{H}}$$

Substitute the given values:

$$m_{{}_{3}^{6}\mathrm{Li}} = 2 \cdot 4.0026 + 0.0239 - 2.014$$

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Step 3: Perform the Calculations

1. Calculate the total mass of alpha particles:

$$2 \cdot 4.0026 = 8.0052 \, \mathrm{u}$$

2. Add the mass contribution from $Q$:

$$8.0052 + 0.0239 = 8.0291 \, \mathrm{u}$$

3. Subtract the mass of the deuteron:

$$8.0291 - 2.014 = 6.0151 \, \mathrm{u}$$

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Answer

The mass of the lithium isotope ${}_{3}\mathrm{Li}^{6}$ is:

$$m_{{}_{3}^{6}\mathrm{Li}} = 6.0151 \, \mathrm{u}$$

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