Question: Given that $M(_{1}^{3}H - atom)$ = 3.016050 amu. $M(_{2}^{3}He - atom)$ = 3.016030 amu Maximum...

Question

Given that $M(_{1}^{3}H - atom)$ = 3.016050 amu.

$M(_{2}^{3}He - atom)$ = 3.016030 amu

Maximum energy of the electron emitted in the b decay of ${}_{1}^{3}H$ is-

Answer 1

Solution

The reaction is

${ } _ { 1 } ^ { 3 } \mathrm { H }$ ® ${ } _ { 2 } ^ { 3 } \mathrm { He }$ + e^– + $E ^ { \prime } = K \left( \frac { 1 } { 1 ^ { 2 } } - \frac { 1 } { \infty ^ { 2 } } \right) = K$

Q = (M_H – M_He)c²

Q = (3.016050 – 3.016030) amu × 931.5 MeV/amu

Q = 0.0186 MeV = K_He + K_e + Kn

M_He >> m_e ; K.E. of helium nucleus will be neglected, so that 0.0186 MeV of energy will be shared between electron and neutrino. When the energy of the neutrino is zero, kinetic energy of the electron will have its maximum value of 0.0186 MeV, i.e. 18.6 KeV

Question: Given that \(M(_{1}^{3}H - atom)\) = 3.016050 amu. \(M(_{2}^{3}He - atom)\) = 3.016030 amu Maximum...

Solution