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Physics Question on Nuclei

In the following reaction, the energy released is 411H→24He+2e++4 ^{1}_{1}H \to ^{4}_{2}He + 2e^{+} + Energy Given : Mass of 11H=1.007825 u^{1}_{1}H = 1.007825 \,u Mass of 24He=4.002603 u^{4}_{2}He = 4.002603\, u Mass of e+=0.000548 ue^{+} = 0.000548\, u

A

12.33 MeV

B

24.67 MeV

C

25.7 MeV

D

49.34 MeV

Answer

25.7 MeV

Explanation

Solution

The given nuclear reaction is 411H→24 He+2e++4_{1} ^{1}H \to ^{4\,}_{2}He + 2e^{+} + Energy The energy released during the process is Q=[4m(1 1H)−m(24 He)−2(me+)]c2Q = \left[4m\left(^{\,1}_{1}H\right) - m\left(^{4\,}_{2}He\right) - 2\left(me_{+}\right)\right]c^{2} =[4?1.007825−4.002603)−2?0.000548]u×c2= \left[4 ? 1.007825 - 4.002603) - 2 ? 0.000548\right]u \times c^{2} =[4.0313−4.002603−0.001096]u?c2= \left[4.0313 - 4.002603 - 0.001096\right]u ? c^{2} =(0.027601 u)c2=(0.027601 u)(931.5 MeV/u)=25.7 MeV= \left(0.027601 \,u\right)c^{2} = \left(0.027601 \,u\right)\left(931.5 \,MeV/u\right) = 25.7\, MeV