An exoergic reaction Now let’s use Equation 30.12 to calculate the reaction energy when lithium is bombarded by a proton and two alpha particles are produced.

Question

An exoergic reaction

Now let’s use Equation 30.12 to calculate the reaction energy when lithium is bombarded by a proton and two alpha particles are produced.

[latex]\mathrm{Q = (M_A+M_B-M_C-M_D)}c^2. [/latex] (30.12)

Accepted Answer

SET UP The reaction can be represented as

[latex]\mathrm{^1_1 H + ^7_3Li \rightarrow {}^4_2He + ^4_2He.}[/latex]

We need to use the definition of reaction energy, Equation 30.12. The masses we need are given in Table 30.2.

SOLVE From Table 30.2, we find the initial and final masses:

[latex]\mathrm{\begin{matrix} A: {}^1_1H & 1.007825 u & C:^4_2He& 4.002603 u \end{matrix} } [/latex]

[latex]\mathrm{\begin{matrix} B: {^7_3Li} &\frac{7.016005 u}{8.023830 u} & D: {}^4_2He &\frac{4.002603 u}{8.005206 u} \end{matrix} } [/latex]

Four electron masses are included on each side. We see that

[latex]\mathrm{M_A + M_B - M_C - M_D = 0.018624 u.}[/latex]

The mass decreases by 0.018624 u; from Equation 30.12, the reaction energy is

[latex]\mathrm{Q = (M_A+M_B-M_C-M_D)}c^2. [/latex] (30.12)

Q = (0.018624 u)(931.5 MeV/u) = 17.35 MeV.

REFLECT The total mass decreases, so the total kinetic energy increases. The final total kinetic energy of the two separating α particles is 17.35 MeV greater than the initial total kinetic energy of the proton and the lithium nucleus.

Practice Problem: Find the Q value for the reaction [latex]\mathrm{^1_0n + ^{10}_5B \rightarrow ^7_3Li + ^4_2He.}[/latex] Is this reaction exoergic or endoergic? Answers: 2.79 MeV, exoergic.

Question 30.7: An exoergic reaction Now let’s use Equation 30.12 to calcula...

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