Consider the following reaction:
`""_(20)Ca^(42) to ""_(20)Ca^(41) + ""_(0)n^(1)`
Calculate the energy E required to remove this neutron from `""_(20)Ca^(42)`.
Further, energy required to remove a proton from `""_(20)Ca^(42)` is E'
What would you suggest whether E' would be less than or greater than E and why?
Mass of `""_(20)Ca^(42) = 41.9586`
Mass of `""_(20)Ca^(41) = 40.9622` u
Mass of neutron =1.0072 u

Explanation

To solve the problem, we need to calculate the energy required to remove a neutron from Calcium-42 (\( _{20}^{42}\text{Ca} \)). This involves calculating the mass defect and then converting that into energy using Einstein's equation \( E = \Delta m c^2 \).

### Step-by-Step Solution:

1. **Identify the masses involved in the reaction**:
- Mass of \( _{20}^{42}\text{Ca} \) (reactant): \( m_{\text{Ca-42}} = 41.9586 \) u
- Mass of \( _{20}^{41}\text{Ca} \) (product): \( m_{\text{Ca-41}} = 40.9622 \) u
- Mass of neutron: \( m_n = 1.0072 \) u
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