Question

Mass of ${}_{4}^{7}Be$atom = 7.016929 amu

Mass of ${}_{3}^{7}Li$atom = 7.016004 amu

${}_{4}^{7}Be$undergoes electron capture at rest. The energy of neutrons is-

• A. 0.2 MeV
• B. 0.9 MeV
• C. 20 MeV
• D. 28.6 MeV

Answer 1

Answer: (B)

0.9 MeV

Answer 2

The electron capture reaction is

${ } _ { 4 } ^ { 7 } \mathrm { Be }$ + e^– ® ${ } _ { 3 } ^ { 7 } \mathrm { Li }$ + n

Q = [M_Be – M_Li] C²

= [7.016929 – 7.016004] × 931.5

= 0.862 MeV

Q = K.E. (Li) + E_n

Because ${ } _ { 3 } ^ { 7 } \mathrm { Li }$ large mass and neutrino has zero rest mass, almost all the energy is carried by neutrino

So that

E_n 0.862 MeV

E_n 0.9 MeV