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Chapter 12

Q. 12.17

Consider the γ decay from the 0.072-MeV excited state to the ground state of { }^{226} Th 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).

h f=E_{>}-E_{<} (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 { }^{226} Th 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, h / \lambda.

p=\frac{h}{\lambda}=\frac{h f}{c}

The conservation of energy gives

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

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


Verified Solution

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


h f+\frac{(h f)^{2}}{2 M c^{2}}=E_{>}-E_{<} (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


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


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


x=\frac{h f}{2 M c^{2}}=\frac{0.072 MeV }{2\left(226 u \cdot c^{2}\right)\left(\frac{931.5 MeV }{c^{2} \cdot u }\right)}=1.7 \times 10^{-7}


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