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In the nuclear fission reaction n + uranium-235 producing cesium-137, rubidium-97, and 2 neutrons, the rest masses are 1.009 u for a neutron, 235.044 u for uranium-235, 136.907 u for cesium-137, and 96.937 u for rubidium-97. Using 1 u = 1.66 x 10^-27 kg and c = 3.00 x 10^8 m/s, what is the energy E released?
A1.8 x 10^-11 J
B2.9 x 10^-11 J
C1.8 x 10^-10 J
D2.9 x 10^-10 J
Answer & Solution
Correct answer: B. 2.9 x 10^-11 J
1. The mass deficit is the uranium-235 mass minus the combined masses of the products: 235.044 minus (136.907 + 96.937 + 1.009) u.
2. Adding the products gives 234.853 u, so the deficit is 235.044 minus 234.853, which is 0.191 u.
3. Convert to kilograms: (0.191 u)(1.66 x 10^-27 kg/u) = 3.17 x 10^-28 kg.
4. Apply E = m c^2: (3.17 x 10^-28 kg)(3.00 x 10^8 m/s)^2.
5. Multiplying gives E of about 2.85 x 10^-11 J, which rounds to 2.9 x 10^-11 J.
6. 1.8 x 10^-11 J and 1.8 x 10^-10 J come from a smaller mass deficit, and 2.9 x 10^-10 J shifts the decimal point one place too far.
7. So 2.9 x 10^-11 J is correct.
_Source: OpenStax Physics (CC BY 4.0), Ch 10 "Special Relativity", section 10.2 Consequences of Special Relativity_
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