In the fusion reaction $_{1}^{2} H +_{1}^{2} H {\to_{}^{}}_{2}^{3} H e +_{0}^{1} n,$ the masses of deuteron helium and neutron expressed in amu are 2.015, 3.017 and 1.009 respectively. If 1kg of deuterium undergoes complete fusion, the amount of total energy released (1amu = 931.5 $M e V / c^{2}$ ) is $p \times 10^{13} J$ . Find $p$ .
(Round off to nearest integer)

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 9.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$Δ m = 2 (2.015) - (3.017 + 1.009) = 0.004 a m u$

Energy released = $(0.004 \times 931.5) M e V = 3.726 M e V$

Number of deuterons in $1 k g = \frac{6.02 \times 10^{26}}{2} = 3.01 \times 10^{26}$

Energy released per kg of deuterium fusion = $(3.01 \times 10^{26} \times 1.863) = 5.6 \times 10^{26} M e V = 9.0 \times 10^{13} J$

Watch 3-min video & get full concept clarity