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If in a nucle...

Question

If in a nuclear fusion process, the masses of the fusion nuclei be $m_{1}$ and $m_{2}$ and the mass of the resultant nucleus be $m_{3}$ , then

$m_{3} = m_{1} + m_{2}$

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$m_{3} = m_{1} - m_{2}$

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$m_{3} < m_{1} + m_{2}$

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$m_{3} > m_{1} + m_{2}$

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Solution

The correct option is C
$m_{3} < m_{1} + m_{2}$

Step 1: Given Data:
Masses of two light nuclei are $m_{1}$ and $m_{2}$ .
Mass of the heavier nucleus produced in the fusion is $m_{3}$ .
Step 2: Nuclear fusion:
In nuclear fusion, two lighter nuclei combine together to produce a heavier nucleus. The process is exoergic, i.e., net energy gets released in the process.
The mass of the heavier nucleus is less than the sum of the mass of the two light nuclei. This mass converts into energy which is released in the fusion process.
Therefore, $m_{3} < m_{1} + m_{2}$
Hence, the correct option is (C).

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Similar questions

Two light nuclei of masses $m_{1}$ and $m_{2}$ are fused to form a more stable nucleus of mass $m_{3}$ then :-

_{Z}^{A} X

(A - Z)

M

m_{1}

m_{2}

(m_{1} + m_{2}) < M .

m_{3}

m_{4}

M^{'}

(m_{3} + m_{4}) > M^{'}

\begin{matrix} _{1}^{1} H & 1.007825 u & _{1}^{2} H & 2.014102 u_{3}^{6} L i & 6.015123 u & _{3}^{7} L i & 7.016004 u_{64}^{152} G d & 151.919803 u & _{82}^{206} P b & 205.974455 u_{1}^{3} H & 3.016050 u & _{1}^{4} H e & 4.002603 u_{30}^{70} Z n & 69.925325 u & _{34}^{82} S e & 81.916709 u_{83}^{209} B i & 208.980388 u & _{84}^{210} P o & 209.982876 u \end{matrix}

(1 u = 932 \frac{M e V}{c^{2}})

m_{p}

m_{n}

M_{1}

_{10} {Ne}^{20}

M_{2}

_{20} {Ca}^{40}

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