Question

If 200 MeV energy is released in the fission of a single nucleus of \({}^{235}_{92}U\), how many fissions per second must occur to produce a power of 1 kW?

• A. \(6.250 \times 10^{15}\)
• B. \(3.125 \times 10^{13}\)
• C. \(4.251 \times 10^{13}\)
• D. \(2.155 \times 10^{14}\)

Hint:

Fissions per second \(= P/E\) with \(E = 200\) MeV converted to joules.

Answer 1

The Correct Option is B

Step 1: Energy per second needed:
A power of 1 kW means the reactor must give out 1000 J every second.

Step 2: Energy from one fission in joules:
One eV is $1.6 \times 10^{-19}$ J. So 200 MeV is $200 \times 10^{6} \times 1.6 \times 10^{-19} = 320 \times 10^{-13} = 3.2 \times 10^{-11}$ J.

Step 3: Count fissions:
Number of fissions in one second is the total energy divided by the energy of one fission.
\[ N = \frac{1000}{3.2 \times 10^{-11}} = \frac{10^{3}}{3.2} \times 10^{11} \]
Since $10^3 / 3.2 = 312.5$, we get $N = 312.5 \times 10^{11} = 3.125 \times 10^{13}$.

Step 4: Match:
This equals the second option.

Final Answer:
The rate is $3.125 \times 10^{13}$ per second. \[\boxed{3.125 \times 10^{13}}\]