Nuclear Reactions Questions and Answers

Q: The following nuclear reaction is induced by bombarding neutrons on the \(^{13}\)C target.\(^{13}\)C + n \(\rightarrow\) \(^{10}\)Be + \(^{4}\)HeIf the R value of this reaction is - 3.835 MeV, the minimum neutron energy required to induce the reaction is nearly :

The minimum neutron energy is nearly 4.13 MeV.For an endothermic reaction the threshold kinetic energy of the projectile exceeds |Q| because momentum must be conserved, so part of the energy goes into motion of the centre of mass:\[ \begin{aligned} E_{th} &= -Q\left(1 + \frac{m_{n}}{M_{C}}\right) \\ &= 3.835 \times \frac{14}{13} \approx 4.13\ \text{MeV} \end{aligned} \]The option 4.835 MeV adds 1 MeV for no reason, and 5.835 MeV adds 2 MeV; 2.13 MeV is below |Q| and could not drive the reaction at all.

Q: To penetrate the Coulomb barrier of a light nucleus (\({}^{1}_{1}H\)), a proton must have a minumum energy of the order of (Consider radius of nucleus ~ 1Fm) :

The proton needs an energy of the order of 1.44 MeV.The Coulomb barrier is the electrostatic potential energy of the two charges when they touch, here at a separation of about 1 fm:\[ \begin{aligned} U &= \frac{e^{2}}{4\pi\varepsilon_{0}r} = \frac{1.44\ \text{MeV fm}}{1\ \text{fm}} \\ &= 1.44\ \text{MeV} \end{aligned} \]using the convenient constant \(e^{2}/4\pi\varepsilon_{0} = 1.44\) MeV fm. Nuclear barriers are of the order of MeV, which is why nuclear reactions need MeV particles; TeV and GeV are far too large, and keV far too small, although quantum tunnelling lets fusion in stars proceed at keV thermal energies.

Q: What is the energy released in the following nuclear reaction :\[ {}^{6}_{3}Li + {}^{2}_{1}H \rightarrow 2\,{}^{4}_{2}He + Q \](Given : \({}^{6}_{3}Li = 6.01693\) amu, \({}^{2}_{1}H = 2.01471\) amu, \({}^{4}_{2}He = 4.00388\) amu)

On the key, the energy released is about 22.4 MeV.The Q value is the mass lost in the reaction:\[ \begin{aligned} \Delta m &= (6.01693 + 2.01471) - 2(4.00388) \\ &= 8.03164 - 8.00776 = 0.02388\ \text{amu} \\ Q &= 0.02388 \times 931.5 = 22.2\ \text{MeV} \end{aligned} \]The keyed 22.4 MeV is the nearest option to this value; the other options are about 11 MeV away. The positive Q shows the reaction releases energy, because the two alpha particles are much more tightly bound than lithium-6 and the deuteron.

Q: The Q value of \({}^{197}_{79}\mathrm{Au} + {}^{12}_{6}\mathrm{C}\) \(\to {}^{206}_{85}\mathrm{At} + ?\) in MeV is :(Given : 1 u = 931.5 MeV/c2, M(\({}^{1}_{1}\mathrm{H}\)) = 1.007825 u, Mn = 1.008665 u, M(\({}^{12}_{6}\mathrm{C}\)) = 12.000000 u, M(\({}^{197}_{79}\mathrm{Au}\)) = 196.966543 u, M(\({}^{206}_{85}\mathrm{At}\)) = 205.986667 u)

The Q value is −43 MeV.Balance the reaction first. The left side has \(A = 197 + 12 = 209\) and \(Z = 79 + 6 = 85\); astatine-206 has \(A = 206\), \(Z = 85\). The missing products carry \(A = 3\), \(Z = 0\): three neutrons. Then\[ \begin{aligned} Q &= [M_{Au} + M_{C} - M_{At} - 3m_{n}]c^{2} \\ &= [208.966543 - 205.986667 - 3.025995]\ \text{u} \\ &= -0.046119\ \text{u} \times 931.5 \approx -43\ \text{MeV} \end{aligned} \]The reaction is endothermic: the beam must supply at least 43 MeV (more, in the laboratory frame).+43 MeV gets the sign backwards. −30 and 30 MeV do not come from these masses.

The following nuclear reaction is induced by bombarding neutrons on the \(^{13}\)C target.

\(^{13}\)C + n \(\rightarrow\) \(^{10}\)Be + \(^{4}\)He

If the R value of this reaction is - 3.835 MeV, the minimum neutron energy required to induce the reaction is nearly :

A.2.13 MeV
B.4.13 MeV
C.4.835 MeV
D.5.835 MeV

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The minimum neutron energy is nearly 4.13 MeV.

For an endothermic reaction the threshold kinetic energy of the projectile exceeds |Q| because momentum must be conserved, so part of the energy goes into motion of the centre of mass:

\[ \begin{aligned} E_{th} &= -Q\left(1 + \frac{m_{n}}{M_{C}}\right) \\ &= 3.835 \times \frac{14}{13} \approx 4.13\ \text{MeV} \end{aligned} \]

The option 4.835 MeV adds 1 MeV for no reason, and 5.835 MeV adds 2 MeV; 2.13 MeV is below |Q| and could not drive the reaction at all.

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To penetrate the Coulomb barrier of a light nucleus (\({}^{1}_{1}H\)), a proton must have a minumum energy of the order of (Consider radius of nucleus ~ 1Fm) :

A.1.44 TeV
B.1.44 GeV
C.1.44 MeV
D.1.44 KeV

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The proton needs an energy of the order of 1.44 MeV.

The Coulomb barrier is the electrostatic potential energy of the two charges when they touch, here at a separation of about 1 fm:

\[ \begin{aligned} U &= \frac{e^{2}}{4\pi\varepsilon_{0}r} = \frac{1.44\ \text{MeV fm}}{1\ \text{fm}} \\ &= 1.44\ \text{MeV} \end{aligned} \]

using the convenient constant \(e^{2}/4\pi\varepsilon_{0} = 1.44\) MeV fm. Nuclear barriers are of the order of MeV, which is why nuclear reactions need MeV particles; TeV and GeV are far too large, and keV far too small, although quantum tunnelling lets fusion in stars proceed at keV thermal energies.

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What is the energy released in the following nuclear reaction :

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

(Given : \({}^{6}_{3}Li = 6.01693\) amu, \({}^{2}_{1}H = 2.01471\) amu, \({}^{4}_{2}He = 4.00388\) amu)

A.11.4 MeV
B.22.4 MeV
C.33.4 MeV
D.44.4 MeV

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

On the key, the energy released is about 22.4 MeV.

The Q value is the mass lost in the reaction:

\[ \begin{aligned} \Delta m &= (6.01693 + 2.01471) - 2(4.00388) \\ &= 8.03164 - 8.00776 = 0.02388\ \text{amu} \\ Q &= 0.02388 \times 931.5 = 22.2\ \text{MeV} \end{aligned} \]

The keyed 22.4 MeV is the nearest option to this value; the other options are about 11 MeV away. The positive Q shows the reaction releases energy, because the two alpha particles are much more tightly bound than lithium-6 and the deuteron.

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The Q value of \({}^{197}_{79}\mathrm{Au} + {}^{12}_{6}\mathrm{C}\) \(\to {}^{206}_{85}\mathrm{At} + ?\) in MeV is :

(Given : 1 u = 931.5 MeV/c², M(\({}^{1}_{1}\mathrm{H}\)) = 1.007825 u, Mn = 1.008665 u, M(\({}^{12}_{6}\mathrm{C}\)) = 12.000000 u, M(\({}^{197}_{79}\mathrm{Au}\)) = 196.966543 u, M(\({}^{206}_{85}\mathrm{At}\)) = 205.986667 u)

A.−30
B.30
C.−43
D.43

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The Q value is −43 MeV.

Balance the reaction first. The left side has \(A = 197 + 12 = 209\) and \(Z = 79 + 6 = 85\); astatine-206 has \(A = 206\), \(Z = 85\). The missing products carry \(A = 3\), \(Z = 0\): three neutrons. Then

\[ \begin{aligned} Q &= [M_{Au} + M_{C} - M_{At} - 3m_{n}]c^{2} \\ &= [208.966543 - 205.986667 - 3.025995]\ \text{u} \\ &= -0.046119\ \text{u} \times 931.5 \approx -43\ \text{MeV} \end{aligned} \]

The reaction is endothermic: the beam must supply at least 43 MeV (more, in the laboratory frame).

+43 MeV gets the sign backwards. −30 and 30 MeV do not come from these masses.

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