Question

The binding energies per nucleon of deuteron ${(}_1{H}^2)$ and helium atom ${(}_{2}H{e}^4)$ are $1.1MeV$ and $7MeV$. If two deuteron atoms react to form a single helium atom, then the energy released is

• A. $13.9MeV$
• B. $26.9MeV$
• C. $23.6MeV$
• D. $19.2MeV$

Answer 1

The correct option is C $23.6MeV$

ATQ, following reaction takes place,

$H_1^2+H_1^2\to H_2^4-Q$

{-Q: energy reloaded}

$\to 2\left(H_1^2\right)\to H_2^4-Q$

So, ATQ binding energy per nucleon of deuteron=1.1MeV.

Therefore, There are 2 nucleon in deuteron atom, hence total binding energy

$=1.1\times 2=2.2MeV$

Therefore, for left hand part, BE of

$H_1^2=2.2MeV$

Similarly, BE per nucleon of helium atom= 7eV.

Number of nucleons in helium$=4$

Total BE of nucleus$==7\times 4=28MeV$

Now, substituting the values of BE in the equation to get the value of Q,

$\Rightarrow 2\left(2.2\right)\to 28-Q\to Q=28-4.4=23.6MeV$