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## Q. 13.1

J. L. Black and colleagues* measured the total cross section for the ${ }^{12} C (\alpha, n)^{15} O$ reaction at an incident energy of $E_{\alpha}=14.6$ MeV to be $\sigma_{T}=25$ mb. If a $1.0-\mu A$ α-particle beam $\left({ }^{4} He ^{++}\right)$ is incident on a 4.0-mm^2 carbon target of thickness $1.0\mu A$ (density $=1.9 g / cm ^{3}$) for one hour, how many neutrons are produced?

Strategy In order to find the number $N_{n}$ of neutrons produced, we find the probability of scattering and multiply it by the number $N_{I}$ of incident α particles: $N_{n}=N_{I} P$. To find the probability of scattering, we first need to determine n, the number of nuclei/volume, which we established in Equation (4.8):

$n=\frac{\rho N_{ A } N_{M}}{M_{g}} \frac{\text { atoms }}{ cm ^{3}}$ (4.8)

We can calculate the probability of scattering $P=n t \sigma$ [Equation (13.5)], because we can determine n and thickness t, and we are given the cross section σ. Lastly, we determine the number $N_{I}$ of incident α particles from the beam current and time the beam is on the target.

$\text { Probability of scattering }=\frac{N_{s} \sigma}{A}=\frac{n t A \sigma}{A}=n t \sigma$ (13.5)

## Verified Solution

Some of the values needed are

$\rho=1.9 g / cm ^{3} \quad N_{ A }=6.02 \times 10^{23} \text { molecules } / mol$

$N_{M}=1 \text { atom/molecule } \quad M_{g}=12 g / mol$

If we substitute these values in the equation for n, we have

\begin{aligned}n=&\left(\frac{1.9 g }{ cm ^{3}}\right)\left(\frac{6.02 \times 10^{23} \text { molecules }}{ mol }\right) \\& \times\left(\frac{1 \text { atom }}{ molecule }\right)\left(\frac{ mol }{12 g }\right) \\=& 9.53 \times 10^{22} \frac{\text { atoms }}{ cm ^{3}}\end{aligned}

The probability of scattering can now be determined from Equation (13.5).

\begin{aligned}P=n t \sigma=&\left(9.53 \times 10^{22} \text { nuclei } / cm ^{3}\right)\left(1.0 \times 10^{-6} m \right) \\& \times\left(25 \times 10^{-31} m ^{2}\right)\left(10^{6} cm ^{3} / m ^{3}\right) \\=& 2.4 \times 10^{-7}\end{aligned}

The number of incident $\alpha \text { particles } N_{I}$ on the target can be determined by the beam current and length of time the beam is on the target.

\begin{aligned}N_{I}=&(1.0 \mu A )\left(\frac{10^{-6} C / s }{\mu A }\right)(1.0 h )\left(3600 \frac{ s }{ h }\right) \\& \times\left[\frac{1 \text { alpha } }{2\left(1.6 \times 10^{-19} C \right)}\right] \\=& 1.1 \times 10^{16} \text { alphas }\end{aligned}

Note that we have taken the charge of the incident α particles to be +2e.

The ratio of detected neutrons to incident alpha particles $\left(N_{n} / N_{I}\right)$ is the probability P of scattering. We therefore have

\begin{aligned}N_{n} &=N_{I} P=N_{I} n t \sigma \\&=\left(1.1 \times 10^{16} \text { alphas }\right)\left(2.4 \times 10^{-7}\right. \text { neutrons/alpha) }\\&=2.6 \times 10^{9} \text { neutrons }\end{aligned}

*J. L. Black et al., Nuclear Physics 115, 683 (1968).