Define the multiplication factor $(k)$ in the context of a nuclear chain reaction.
(N/A) The multiplication factor $(k)$ is defined as the ratio of the number of neutrons present at the beginning of a particular generation to the number of neutrons present at the beginning of the preceding generation.
Mathematically,$k = \frac{\text{Number of neutrons in a given generation}}{\text{Number of neutrons in the preceding generation}}$.
- If $k = 1$,the chain reaction is steady (critical state).
- If $k > 1$,the chain reaction is accelerating (supercritical state).
- If $k < 1$,the chain reaction dies out (subcritical state).
Mathematically,$k = \frac{\text{Number of neutrons in a given generation}}{\text{Number of neutrons in the preceding generation}}$.
- If $k = 1$,the chain reaction is steady (critical state).
- If $k > 1$,the chain reaction is accelerating (supercritical state).
- If $k < 1$,the chain reaction dies out (subcritical state).
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$_1H^1 + _1H^1 + _1H^2 \to X + _1e^0 + \text{Energy}$. The emitted particle $X$ is ........?
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Given atomic masses of:
${ }^{235} U: 235.0439 \text{ u}, { }^{140} Ce: 139.9054 \text{ u},$
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Given atomic masses of:
${ }^{235} U: 235.0439 \text{ u}, { }^{140} Ce: 139.9054 \text{ u},$
${ }^{94} Zr: 93.9063 \text{ u}, n: 1.0086 \text{ u},$
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View Solution$A$ star has $10^{40}$ deuterons. It produces energy via the processes:
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If the average power radiated by the star is $10^{16} \ W$, the deuteron supply of the star is exhausted in a time of the order of:
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$_1H^2 + _1H^3 \to _2He^4 + n$
If the average power radiated by the star is $10^{16} \ W$, the deuteron supply of the star is exhausted in a time of the order of:
Given:
Mass of $_1H^2 = 2.014 \ amu$
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View SolutionTwo lighter nuclei combine to form a comparatively heavier nucleus by the relation given below:
${ }_{1}^{2} X +{ }_{1}^{2} X ={ }_{2}^{4} Y$
The binding energies per nucleon of ${ }_{1}^{2} X$ and ${ }_{2}^{4} Y$ are $1.1 \, MeV$ and $7.6 \, MeV$ respectively. The energy released in this process is in $MeV$.
${ }_{1}^{2} X +{ }_{1}^{2} X ={ }_{2}^{4} Y$
The binding energies per nucleon of ${ }_{1}^{2} X$ and ${ }_{2}^{4} Y$ are $1.1 \, MeV$ and $7.6 \, MeV$ respectively. The energy released in this process is in $MeV$.
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