Question

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

Answer 1

Correct Answer: 2.33

\(^{16}_{7}\text{N} + ^{4}_{2}\text{He} \rightarrow ^{1}_{1}\text{H} + ^{19}_{8}\text{O}\)

4v₀ = 1v₁ + 19v₂ = 20v₂ (For max loss of KE)

\(v_0=\frac{v_2}{5}\)

E required \(= (1.008 + 19.003 – 16.006 – 4.003) × 930 = 1.86\)

\(\frac{1}{2}4v_0^2-\frac{1}{2}20v^2=1.86\)

\(\frac{1}{2}4v_0^2-10\frac{v_0^2}{25}20v^2=1.86\)

\(2v_0^2-\frac{2}{5}v_0^2=1.86\)

\(v_0^2=\frac{1.86\times5}{8}\)

KE=\(\frac{1}{2}4v_0^2=2v_0^2=\frac{18.6\times5}{4}\)

\(=2.325\)

Answer 2

To determine the minimum kinetic energy needed by an alpha particle to cause the nuclear reaction:

\[ ^{16}_{7}\text{N} + ^{4}_{2}\text{He} \rightarrow ^{1}_{1}\text{H} + ^{19}_{8}\text{O} \]

we can follow this detailed analysis:

1. Reaction Q-value Calculation
  Given the masses:

\( m(^{16}_{7}\text{N}) = 16.006 \, \text{u} \)
\( m(^{4}_{2}\text{He}) = 4.003 \, \text{u} \)
\( m(^{1}_{1}\text{H}) = 1.008 \, \text{u} \)
\( m(^{19}_{8}\text{O}) = 19.003 \, \text{u} \)
\( 1 \, \text{u} = 930 \, \text{MeV}/c^2 \)

  The Q-value for the reaction is:
\[Q = (m_{\text{reactants}} - m_{\text{products}}) \times c^2 \]
\[ Q = (16.006 \, \text{u} + 4.003 \, \text{u} - 1.008 \, \text{u} - 19.003 \, \text{u}) \times 930 \, \text{MeV/u} \]
\[ Q = (20.009 \, \text{u} - 20.011 \, \text{u}) \times 930 \, \text{MeV/u}\]
\[ Q = -0.002 \, \text{u} \times 930 \, \text{MeV/u} \]
\[ Q = -1.86 \, \text{MeV} \]
  Since \( Q \) is negative, 1.86 MeV of energy must be supplied for the reaction to occur.

2. Kinetic Energy Calculation:
  Using momentum conservation for maximum energy loss:

  \[ 4v_0 = 1v_1 + 19v_2 = 20v_2 \quad \text{(for max loss of KE)} \]
  \[ v_0 = \frac{v_2}{5} \]
  The kinetic energy required:
  \[ \frac{1}{2}4v_0^2 - \frac{1}{2}20v_2^2 = 1.86 \]
  Substituting \( v_0 = \frac{v_2}{5} \):
  \[  \frac{1}{2} \left( 4 \left( \frac{v_2}{5} \right)^2 \right) - \frac{1}{2} \left( 20v_2^2 \right) = 1.86  \]
  \[\frac{1}{2} \left( 4 \times \frac{v_2^2}{25} \right) - 10v_2^2 = 1.86\]
  \[\frac{2v_2^2}{25} - 10v_2^2 = 1.86 \]
  \[  2v_2^2 \left( \frac{1}{25} - \frac{5}{25} \right) = 1.86  \]
  \[  2v_2^2 \left( -\frac{4}{25} \right) = 1.86 \]
  \[ v_2^2 = \frac{1.86 \times 25}{-8} \times -1 = \frac{1.86 \times 25}{8} \]

  Hence:
  \[ v_0^2 = \frac{v_2^2}{25} = \frac{1.86 \times 25}{8 \times 25} = \frac{1.86}{8} \]

  The kinetic energy:
  \[  \text{KE} = 2v_0^2 = 2 \times \frac{1.86 \times 5}{8} = \frac{1.86 \times 10}{8} = \frac{18.6}{8} = 2.325 \, \text{MeV}  \]

 Conclusion

Thus, the minimum kinetic energy needed by the alpha particle for the reaction is:

\[ {2.325 \, \text{MeV}} \]