A neutron breaks into a proton, an electron (ß-particle) and antineutrino . 0 n1 longrightarrow 1 H 1+ -1 β0+ barv). The energy released in the process in MeV is

Question

A neutron breaks into a proton, an electron (ß-particle) and antineutrino . 0 n1 longrightarrow 1 H 1+ -1 β0+ barv). The energy released in the process in MeV is (A) 731 MeV (B) 7.31 MeV (C) 0.731 MeV (D) 1.17 MeV

Tardigrade Admin · Accepted Answer

From Einstein's mass energy relation Δ E=(Δ m) c2 Mass of proton =1.6725 × 10-27 kg Mass of electron =0.0009 × 10-27 kg Their sum =1.6734 × 10-27 kg Mass of neutron =1.6747 × 10-27 kg Their difference Δ m=0.0013 × 10-27 kg ∴ Δ E=(0.0013 × 10-27)(3 × 108)2 =1.17 × 10-13 J Also, 1.6 × 10-19 J =1 eV ∴ Δ E=(1.17 × 10-13/1.6 × 10-19) =0.731 × 106 eV =0.731 MeV .

Q. A neutron breaks into a proton, an electron (ß-particle) and antineutrino 0​n1⟶1​H1+−1​β0+vˉ). The energy released in the process in MeV is

731 MeV

7.31 MeV

0.731 MeV

1.17 MeV

Q. A neutron breaks into a proton, an electron (ß-particle) and antineutrino $_{0} n^{1} ⟶_{1} H^{1} +_{- 1} β^{0} + \overset{v}{ˉ})$ . The energy released in the process in $M e V$ is