Atoms And Nuclei Question 212
Question: 200 MeV of energy may be obtained per fission of $ U^{235} $ . A reactor is generating 1000 kW of power. The rate of nuclear fission in the reactor is [MP PET 1995]
Options:
A) 1000.0
B) $ 2\times 10^{8} $
C) $ 3.125\times 10^{16} $
931
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Answer:
Correct Answer: C
Solution:
Power is the rate at which work is done or energy is transferred. $ \Rightarrow {{( \frac{1}{2} )}^{4}}={{( \frac{1}{2} )}^{t/48}}\Rightarrow t=192\ hour. $ Rate of nuclear fission $ =\frac{10^{6}}{200\times 1.6\times {{10}^{-13}}} $ = 3.125 ´ 1016.
BETA
