The energy released by the fission of one uranium atom is 200 MeV . The number of fissions per second required to produce 3.2 ~W of power is (Take 1 eV=1.6 × 10^-19 ~J )

Correct answer: (D) 10^11. Hints: u=200 MeV=200 × 10^6 eV=200 × 10^6 × 1.6 × 10^-19 ~J E=3.2 ~J No of fissions =frac3.22 × 1.6 × 10^-11=10^11

Q140 / 160 · WBJEE 2010 · Physics

The energy released by the fission of one uranium atom is $200 \mathrm{MeV}$. The number of fissions per second required to produce $3.2 \mathrm{~W}$ of power is (Take $1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$ )

A
$10^7$
B
$10^{10}$
C
$10^{15}$
Correct answer
D
$10^{11}$

Solution

Hints: $u=200 \mathrm{MeV}=200 \times 10^6 \mathrm{eV}=200 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J}$ $\mathrm{E}=3.2 \mathrm{~J}$ No of fissions $=\frac{3.2}{2 \times 1.6 \times 10^{-11}}=10^{11}$

Hence, the correct answer is option: 'D'.

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