Question:
On bombarding $U^{235}$ by slow neutron, $200\, MeV$ energy is released. If the power output of atomic reactor is $1.6\, MW$, then the rate of fission will be
On bombarding $U^{235}$ by slow neutron, $200\, MeV$ energy is released. If the power output of atomic reactor is $1.6\, MW$, then the rate of fission will be
Updated On: Apr 2, 2024
- $5\times10 ^{22}/s$
- $5\times10 ^{16}/s$
- $8\times10 ^{16}/s$
- $20\times10 ^{16}/s$
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The Correct Option is B
Solution and Explanation
Energy released on bombarding $U ^{235}$ by neutron
$=200 \,MeV$
Power output of atomic reactor $=1.6 \,MW$
$\therefore$ Rate of fission $=\frac{1.6 \times 10^{6}}{200 \times 10^{6} \times 1.6 \times 10^{-19}}$
$=5 \times 10^{16} / s$
$=200 \,MeV$
Power output of atomic reactor $=1.6 \,MW$
$\therefore$ Rate of fission $=\frac{1.6 \times 10^{6}}{200 \times 10^{6} \times 1.6 \times 10^{-19}}$
$=5 \times 10^{16} / s$
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Concepts Used:
Nuclei
In the year 1911, Rutherford discovered the atomic nucleus along with his associates. It is already known that every atom is manufactured of positive charge and mass in the form of a nucleus that is concentrated at the center of the atom. More than 99.9% of the mass of an atom is located in the nucleus. Additionally, the size of the atom is of the order of 10-10 m and that of the nucleus is of the order of 10-15 m.
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