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Nuclear Physics: Question 10

Syllabus 23.1

Multiple choice A2 1 mark

A nuclear power station's reactor generates a steady thermal power output of 3.00 GW3.00\text{ GW}. Assuming this energy originates entirely from the conversion of mass to energy via E=mc2E=mc^2, and taking c=3.00×108 m s−1c=3.00\times10^8\text{ m s}^{-1}, calculate the rate at which mass is being converted to energy inside the reactor.

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

Step 1: Recall E = mc² in rate form

Power is energy transferred per second, so dividing both sides of E=mc2E=mc^2 by time converts it into a rate equation: dmdt=Pc2\frac{dm}{dt}=\frac{P}{c^2}

Step 2: Substitute the values

The power output is 3.00 GW=3.00×109 W3.00\text{ GW}=3.00\times10^9\text{ W}, and: c2=(3.00×108)2=9.00×1016 m2 s−2c^2=(3.00\times10^8)^2=9.00\times10^{16}\text{ m}^2\text{ s}^{-2}

So: dmdt=3.00×1099.00×1016\frac{dm}{dt}=\frac{3.00\times10^9}{9.00\times10^{16}}

Dividing the mantissas and the powers of ten separately: 3.009.00=0.3333,1091016=10−7\frac{3.00}{9.00}=0.3333,\qquad\frac{10^9}{10^{16}}=10^{-7}

dmdt=0.3333×10−7=3.33×10−8 kg s−1\frac{dm}{dt}=0.3333\times10^{-7}=3.33\times10^{-8}\text{ kg s}^{-1}

Recompute as a check, rewriting 9.00×10169.00\times10^{16} as 3.00×109×3.00×1073.00\times10^{9}\times3.00\times10^{7}: dmdt=13.00×107=3.33×10−8 kg s−1\dfrac{dm}{dt}=\dfrac{1}{3.00\times10^{7}}=3.33\times10^{-8}\text{ kg s}^{-1}, the same result.

dmdt=3.33×10−8 kg s−1 (3 s.f.)\frac{dm}{dt}=\boxed{3.33\times10^{-8}}\text{ kg s}^{-1}\text{ (3 s.f.)}

Why the other options are wrong

  • B (1.67×10−8 kg s−11.67\times10^{-8}\text{ kg s}^{-1}): half the correct value, arises from an extra factor of 2 in the denominator, e.g. mistakenly using dm/dt=P/(2c2)dm/dt=P/(2c^2).
  • C (3.33×10−7 kg s−13.33\times10^{-7}\text{ kg s}^{-1}): ten times too large, arises from a power-of-ten slip when squaring cc, e.g. computing c2c^2 as 9.00×10159.00\times10^{15} instead of 9.00×10169.00\times10^{16}.
  • D (10.0 kg s−110.0\text{ kg s}^{-1}): comes from forgetting to square cc entirely and computing P/c=3.00×109/3.00×108=10.0 kg s−1P/c=3.00\times10^9/3.00\times10^8=10.0\text{ kg s}^{-1}. This is far too large for a nuclear reaction, a useful sanity check that something has gone wrong.

Final answer

  • The rate of mass-to-energy conversion is 3.33×10−8 kg s−1\boxed{3.33\times10^{-8}}\text{ kg s}^{-1}, option A.