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The mass defect of a beryllium-7 nucleus is 0.0404 u. How much energy is needed to break this nucleus into its component protons and neutrons (in MeV)? (A) 20.8 (B) 37.6 (C) 54.4 (D) 71.2 (E) 88.1

The mass defect of a beryllium-7 nucleus is 0.0404 u. How much energy is needed to break this nucleu...
Answer

The energy needed is the nuclear binding energy, given by $E = \Delta m c^2$. Using $1\,\text{u} = 931.5\,\text{MeV}/c^2$, $E = (0.0404\,\text{u})(931.5\,\text{MeV}/\text{u}) \approx 37.6\,\text{MeV}$. So the correct choice is (B) 37.6 MeV.

Explanation

What the mass defect is telling you

A nucleus has less mass than its separated protons and neutrons. That “missing” mass (the mass defect $\Delta m$) corresponds to the energy released when the nucleus formed, and it is exactly the energy you must add to pull it apart (the binding energy).

Convert mass defect to energy

Use Einstein’s relation $E = \Delta m c^2$ and the handy conversion

$$1\,\text{u} = 931.5\,\text{MeV}/c^2.$$

So the binding energy in MeV is

$$E = (\Delta m)(931.5\,\text{MeV}/\text{u}).$$

Plug in the number for beryllium-7

Given $\Delta m = 0.0404\,\text{u}$:

$$E = (0.0404)(931.5)\,\text{MeV} \approx 37.6\,\text{MeV}.$$

Match to the multiple-choice options

$37.6\,\text{MeV}$ corresponds to (B).

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nuclear physics mass-energy equivalence unit conversion binding energy calculations

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