Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The binding energy of deuteron $({}_1{H^2})$ is $1.15MeV$ per nucleon, and an alpha particle $({}_2H{e^4})$ has a binding energy of $7.1MeV$ per nucleon. Then in the reaction, ${}_1{H^2} + {}_1{H^2} \to {}_2H{e^4} + Q$ . The energy $Q$ released in $MeV$is:

Last updated date: 16th Jun 2024
Total views: 52.8k
Views today: 1.52k
Verified
52.8k+ views
Hint: As we know the mass of a deuteron and an alpha particle, we can calculate the equivalent energies of the total masses from the given values of binding energy per nucleon. Then from the given reaction, we can calculate the released energy.

Complete step by step solution:
Binding energy per nucleon for deuteron${}_1{H^2}$ is $1.15MeV$
Now, we know that mass of a ${}_1{H^2}$ is $2.01478amu$
And the mass of an alpha particle $({}_2H{e^4})$ is $4.00388amu$ .
So, total mass of the two deuterons is
$\Rightarrow (2 \times 2.01478)amu$
$\Rightarrow 4.02956amu$
Now, the equivalent energy to the two deuterons $(2.{}_1{H^2})$ is
$\Rightarrow (4.02956 \times 1.5)MeV$
$\Rightarrow 6.04434MeV$
And, the energy equivalent to the alpha particle is
$\Rightarrow (7.1 \times 4.00388)MeV$
$\Rightarrow 28.427MeV$
Now, the given reaction is, ${}_1{H^2} + {}_1{H^2} \to {}_2H{e^4} + Q$
Comparing the energies on both sides of this reaction, we can have,
$\Rightarrow 6.04434MeV = 28.427MeV + Q$
Therefore, the value of energy released is
$\Rightarrow (28.427 - 6.04434)MeV$
$\Rightarrow 22.38MeV$

If $Z$ = atomic number, $A$ = mass number of the nucleus and, ${m_p}$ , ${m_n}$ be the mass of proton and mass of neutron independently, then,
$Z{m_p}{c^2} + (A - Z){m_n}{c^2} = {M_{Z,A}}{c^2} + \Delta E$
Where, ${M_{Z,A}}$ = mass of nucleus
And, $\Delta E$ = binding energy