Show that `._(92)^(230)U` does not decay by emitting a neutron or proton. Given:
`M(._(92)^(230)U)=230.033927 am u, M(._(92)^(230)U)=229.033496 am u`,
`M(._(92)^(229)Pa)=229.032089 am u, M(n)=1.008665 am u`
`m(p)=1.007825 am u`.

Explanation

To show that \( _{92}^{230}U \) does not decay by emitting a neutron or a proton, we need to analyze both potential decay processes and calculate the Q values for each reaction.

### Step 1: Neutron Emission
Assuming the decay of \( _{92}^{230}U \) by emitting a neutron, the reaction can be written as:

\[ _{92}^{230}U \rightarrow _{92}^{229}U + n \]
...