Modern Physics 5 Question 26
29. Consider the reaction: $\quad{ } _1^{2} H+{ } _1^{2} H={ } _2^{4} He+Q$. Mass of the deuterium atom $=2.0141 u$. Mass of helium atom $=4.0024 u$. This is a nuclear …….. reaction in which the energy $Q$ released is …… MeV.
(1996, 2M)
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Answer:
Correct Answer: 29. $3.32 \times 10^{-5} W$
Solution:
- $Q=(\Delta m$ in atomic mass unit $) \times 931.4 MeV$
$=\left(2 \times\right.$ mass of ${ } _1 H^{2}-$ mass of $\left.{ } _2 He^{4}\right) \times 931.4 MeV$
$=(2 \times 2.0141-4.0024) \times 931.4 MeV$
$Q \approx 24 MeV$ (fusion)
