Consider the γ decay from the 0.072-MeV excited state to the ground state of ^226Th at rest shown in Figure 12.16. Find an exact expression for the gamma-ray energy by including both the conservation of momentum and energy. Determine the error obtained by using the approximate value in Equation

Question

Consider the γ decay from the 0.072-MeV excited state to the ground state of [latex]{ }^{226} Th[/latex] at rest shown in Figure 12.16. Find an exact expression for the gamma-ray energy by including both the conservation of momentum and energy. Determine the error obtained by using the approximate value in Equation (12.46).

[latex]h f=E_{>}-E_{<}[/latex] (12.46)

Strategy We need to account for the conservation of momentum as well as that of energy to find the exact gamma-ray energy. We denote the final momentum of [latex]{ }^{226} Th[/latex] by p. Because the decaying nucleus is initially at rest, the total linear momentum is zero, and the linear momentum p of the daughter nucleus must have the same magnitude but opposite direction to the momentum of the gamma ray, [latex]h / \lambda[/latex].

[latex]p=\frac{h}{\lambda}=\frac{h f}{c}[/latex]

The conservation of energy gives

[latex]h f+\frac{p^{2}}{2 M}=E_{>}-E_{<}[/latex]

where M is the mass of [latex]{ }^{226} Th[/latex]. We solve these two equations to find the gamma-ray energy hf.

Accepted Answer

We substitute the relation for the momentum p into the energy equation and find

[latex]h f+\frac{(h f)^{2}}{2 M c^{2}}=E_{>}-E_{<}[/latex] (12.47)

We could solve Equation (12.47) for hf by using the quadratic equation, but that would be tedious. Let us determine whether we can use an approximation. Rewrite Equation (12.47) as

[latex]\begin{array}{r}h f\left(1+\frac{h f}{2 M c^{2}} ight)=E_{>}-E_{<} \h f=\frac{E_{>}-E_{<}}{1+\frac{h f}{2 M c^{2}}}\end{array}[/latex]

If the value [latex]h f / 2 M c^{2}=x[/latex] is very small, then we can use the binomial expansion [latex](1+x)^{-1}=1-x+x^{2}-\cdots[/latex]. We determine x to be

[latex]x=\frac{h f}{2 M c^{2}}=\frac{0.072 MeV }{2\left(226 u \cdot c^{2} ight)\left(\frac{931.5 MeV }{c^{2} \cdot u } ight)}=1.7 imes 10^{-7}[/latex]

So the error in using the approximate Equation (12.46), which amounts to letting [latex](1+x)^{-1} \approx 1[/latex], amounts to an error of only about [latex]10^{-7}[/latex]. We can safely use Equation (12.46) where [latex]h f \approx E_{>}-E_{<}=0.072[/latex] MeV.

Question 12.17: Consider the γ decay from the 0.072-MeV excited state to the...

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