Protons srtiking a stationary lithium target, activate a reaction `Li^(7)`(p,n)`Be^(7)`. At what value of the proton's kinetic energy can the resulting neutron be stationary?

Explanation

To solve the problem, we need to determine the kinetic energy of a proton striking a stationary lithium-7 nucleus such that the resulting neutron is stationary after the reaction. The reaction is given by:

\[ \text{Li}^{7} + p \rightarrow \text{Be}^{7} + n \]

### Step-by-Step Solution:

1. **Understand the Reaction**:
- A proton (\(p\)) collides with a stationary lithium-7 nucleus (\(\text{Li}^{7}\)) to produce a beryllium-7 nucleus (\(\text{Be}^{7}\)) and a neutron (\(n\)).
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