Question

²³Ne decays to ²³Na by negative beta emission. Mass of ²³Ne is 22.994465 amu mass of ²³Na is 22.989768 amu. The maximum kinetic energy of emitted electrons neglecting the kinetic energy of recoiling product nucleus is ______ MeV.

Options

• 4

• 3

• 2

• 1

Answer 1

²³Ne decays to ²³Na by negative beta emission. Mass of ²³Ne is 22.994465 amu mass of ²³Na is 22.989768 amu. The maximum kinetic energy of emitted electrons neglecting the kinetic energy of recoiling product nucleus is 4 MeV.

Explanation:

\[\ce{_10^23Ne -> _11^23Na + e^- + \overline{v}}\]

Q = [m (²³Ne) - m(²³Na)] × 931.5 Me V

A = 4.375 Me V = 4.4 me V

Q ≃ 4 MeV

Q = KE_y + KE_e + E`overline"v"`

KE_y is very very small

A ≈ KE_e + E`overline"v"`

when KE_e is maximum E`overline"v"` is negligible

KE_e ≃ Q = 4 MeV