Modern Physics Ques 200
- 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)
Show Answer
Answer:
Correct Answer: 200.(Fusion , $24$)
Solution:
Formula:
- $Q=(\Delta m$ 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)
BETA
