A reactor is generating 1000 kW of power and 200 MeV of energy may be obtained per fission of U235. The rate of nuclear fission in the reactor is
Answers
Answered by
46
Answer:
Rate of nuclear fission is 3.125×10^16.
Explanation:
Power generated by the nuclear reactor is 1000 W
Energy to be obtained from the reactor is 2 MeV
The energy released in every fission of U235 is given in the question. We know the formula P = nE/t
Hence, substituting the values in this formula, we get
n/t = P/E = 10^6 / 200[1.6×10^-13]
n/t = 3.125×10^16
Hence Rate of nuclear fission is 3.125×10^16.
Answered by
3
Explanation:
The energy released in every fission of U
236
is given in the question.
We have P=n
t
E
Hence, substituting the values in the formula, we get
t
n
=
E
P
=1000×10
3
/200[1.6×10
13
]
=3.125×
16
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