The binding energy of deuteron $(_{1}^{2} H)$ is 1.15 MeV per nucleon and an alpha particle $(_{2}^{4} H e)$ has a binding energy of 7.1 MeV per nucleon. Then in the reaction
$_{1}^{2} H +_{1}^{2} H \to_{2}^{4} H e + Q$
the energy Q released is

Q. Calculate the energy released in joules and

M e V

in the following nuclear reaction.

_{1}^{2} H +_{1}^{2} H \to_{2}^{3} H e +_{0}^{1} n

Assume that the masses of

_{1}^{2} H,_{2}^{3} H e

and neutron

(n)

respectively are

2.0141, 3.0160

and

1.0087

in amu.

Q. Calculate the mass loss in the following:

_{1}^{2} H +_{1}^{3} H ⟶

_{2}^{4} H e +_{0}^{1} n

Given the masses:

_{1}^{2} H = 2.014

a m u

_{1}^{3} H = 3.016

a m u

;

_{2}^{4} H e = 4.004

a m u

_{0}^{1} n = 1.008

a m u

Join BYJU'S Learning Program

In a nuclear reaction,21H+21H⟶ 32He+10nif the masses of 21H and 32He are 2.014741 amu and 3.016977 amu respectively, then the Q-value of the reaction is nearly:

In a nuclear reaction,
$_{1}^{2} H +_{1}^{2} H ⟶$ $_{2}^{3} H e +_{0}^{1} n$
if the masses of $_{1}^{2} H$ and $_{2}^{3} H e$ are $2.014741$ amu and $3.016977$ amu respectively, then the $Q$ -value of the reaction is nearly: