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## 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.