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Question

Physics Question on Nuclear physics

what is the Q-value of the reaction p+7Li4He+4Hep+{ }^{7} Li \rightarrow{ }^{4} He +{ }^{4} He The atomic masses of 1H,4He{ }^{1} H ,{ }^{4} He and 7Li{ }^{7} Li are 1.007825u,4.002603u1.007825\, u , 4.002603\,u and 7.016004u7.016004\, u respectively

A

17.35 MeV

B

18.06 MeV

C

177.35 MeV

D

170.35 MeV

Answer

17.35 MeV

Explanation

Solution

The total energy released in reaction (Q)(Q) is equal to the difference in kinetic energy of the final particles and of initial particles.
The total mass of the initial particles
m1=1.007825+7.016004m_{1}=1.007825+7.016004
=8.023829u=8.023829\, u
and the total mass of final particles
mf=2×4.002603m_{f}=2 \times 4.002603
=8.005206u=8.005206 u
Difference between initial and final mass of particles
Δm=mimf\Delta m=m_{i}-m_{f}
=8.0238298.005206=8.023829-8.005206
=0.018623u=0.018623\, u
The Q-value is given by
Q=(Δm)c2Q=(\Delta m) c^{2}
=0.018623×93.15=0.018623 \times 93.15
=17.35MeV=17.35\, MeV
Difference between initial and final mass of particles
The QQ-value is given by