E = mc²: Turn a Power Into a Mass Loss
Einstein's relation E = m c² links any energy released to a loss of mass. Two questions follow: how much energy is equivalent to a given mass, and how fast a source such as a star loses mass to keep radiating.
Mass-Energy Equivalence is the central idea in this question pattern. Identify the condition in the stem before applying the method below.
Energy equivalent of an atomic mass unit
One atomic mass unit, 1.6605 x 10⁻²⁷ kg, is equivalent to E = (1.6605 x 10⁻²⁷)(2.9979 x 10⁸)² = 1.49 x 10⁻¹⁰ J, which is 931 MeV (dividing by 1.602 x 10⁻¹³ J per MeV). This number, about 931 MeV per u, is used in every nuclear mass-defect calculation.
Mass lost by a star
For flux 5.0 x 10⁻⁸ W/m² from a star 300 light years (2.84 x 10¹⁸ m) away, L is about 5 x 10³⁰ W, and the mass converted to radiation each second is dm/dt = L / c² = 5 x 10³⁰ / (9 x 10¹⁶), about 5.6 x 10¹³ kg per second. Convert light years to metres (1 ly is about 9.46 x 10¹⁵ m) before squaring the distance.
OlympIQ exam lens
- ✓E = m c²; 1 u is equivalent to about 931 MeV
- ✓Luminosity L = 4π d² F from the flux F at distance d
- ✓Mass loss per second = L / c²
- ✓Convert light years to metres before using the inverse-square law
Do not use the distance in light years in the inverse-square formula. The flux is in watts per square metre, so convert the distance to metres before computing the luminosity.
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