The binding energy of deuteron $_1^2H$ is $1.112 \, MeV$ per nucleon and an $\alpha$-particle $_2^4He$ has a binding energy of $7.047 \, MeV$ per nucleon. Then in the fusion reaction $_1^2H + _1^2H \to _2^4He + Q$,the energy $Q$ released is ........ $MeV$.

  • A
    $1$
  • B
    $11.9$
  • C
    $23.8$
  • D
    $931$

Explore More

Similar Questions

Two 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 $MeV$.

Fusion reaction takes place at high temperature because

Fusion reaction takes place at high temperature because

When Uranium is bombarded with neutrons,it undergoes fission. The fission reaction can be written as ${}_{92}U^{235} + {}_0n^1 \to {}_{56}Ba^{141} + {}_{36}Kr^{92} + 3x + Q$,where three particles named $x$ are produced and energy $Q$ is released. What is the name of the particle $x$?

If $200 \text{ MeV}$ of energy is released in the fission of one nucleus of ${ }_{92}^{236} U$, the number of nuclei that must undergo fission to release an energy of $1000 \text{ J}$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo