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Example · Example 3

Q.Define the atomic mass unit (u). Given that 1 u=1.660539×10−27 kg1\ \text{u} = 1.660539\times10^{-27}\ \text{kg}, show that its energy equivalent is close to 931.5 MeV931.5\ \text{MeV}.

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✓ Free question

The atomic mass unit is defined as exactly one-twelfth of the mass of a 612C^{12}_{6}\text{C} atom, 1 u=1.660539×10−27 kg1\ \text{u}=1.660539\times10^{-27}\ \text{kg}. By Einstein's relation E=mc2E=mc^2 (Section 8.6), its energy equivalent is

E=mc2=(1.660539×10−27 kg)×(3×108 m/s)2≈1.494×10−10 JE = mc^2 = (1.660539\times10^{-27}\ \text{kg})\times(3\times10^8\ \text{m/s})^2 \approx 1.494\times10^{-10}\ \text{J}

Converting to electron-volts using 1 MeV=1.602×10−13 J1\ \text{MeV} = 1.602\times10^{-13}\ \text{J}:

E=1.494×10−101.602×10−13 MeV≈931.5 MeVE = \frac{1.494\times10^{-10}}{1.602\times10^{-13}}\ \text{MeV} \approx 931.5\ \text{MeV}

which confirms the standard conversion used throughout nuclear physics.

✓Final answer

1 u≈931.5 MeV1\ \text{u} \approx 931.5\ \text{MeV}.

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