Problem

Determine the kinetic energy ($K.E.$) required to accelerate a proton to induce the nuclear reaction:

$${}_{3}^{7}\mathrm{Li} + {}_{1}^{1}\mathrm{H} \rightarrow {}_{4}^{7}\mathrm{Be} + {}_{0}^{1}\mathrm{n} + Q$$

Data

• Mass of proton (${}_{1}^{1}\mathrm{H}$): $1.00814 \, \mathrm{u}$
• Mass of lithium isotope (${}_{3}^{7}\mathrm{Li}$): $7.01823 \, \mathrm{u}$
• Mass of beryllium isotope (${}_{4}^{7}\mathrm{Be}$): $7.01592 \, \mathrm{u}$
• Mass of neutron (${}_{0}^{1}\mathrm{n}$): $1.00866 \, \mathrm{u}$
• Conversion factor:
  • $1 \, \mathrm{u} = 931.5 \, \mathrm{MeV}$

Prerequisite Concepts

1. Energy Released ($Q$):
   The energy released during the reaction is calculated from the mass defect ($\Delta m$) as:

$$Q = \Delta m \cdot 931.5 \, \mathrm{MeV}$$

Where:

$$\Delta m = \left( m_{{}_{1}^{1}\mathrm{H}} + m_{{}_{3}^{7}\mathrm{Li}} \right) - \left( m_{{}_{4}^{7}\mathrm{Be}} + m_{{}_{0}^{1}\mathrm{n}} \right)$$

2. Mass-Energy Equivalence:
   The difference in mass directly corresponds to the energy released in the reaction.

Solution

Step 1: Calculate the Mass Defect ($\Delta m$)

Using the formula:

$$\Delta m = \left( m_{{}_{1}^{1}\mathrm{H}} + m_{{}_{3}^{7}\mathrm{Li}} \right) - \left( m_{{}_{4}^{7}\mathrm{Be}} + m_{{}_{0}^{1}\mathrm{n}} \right)$$

Substitute the given values:

$$\Delta m = \left( 1.00814 + 7.01823 \right) - \left( 7.01592 + 1.00866 \right)$$

Perform the calculation:

$$\Delta m = 8.02637 - 8.02458$$ $$\Delta m = 0.00179 \, \mathrm{u}$$

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Step 2: Convert Mass Defect to Energy ($Q$)

Using the formula:

$$Q = \Delta m \cdot 931.5 \, \mathrm{MeV}$$

Substitute $\Delta m = 0.00179 \, \mathrm{u}$:

$$Q = 0.00179 \cdot 931.5$$

Perform the calculation:

$$Q = 1.67 \, \mathrm{MeV}$$

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Answer

The kinetic energy ($K.E.$) required to induce the nuclear reaction is:

$$Q = 1.67 \, \mathrm{MeV}$$

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