Calculate the kinetic energy of `beta`-particles and the radiation frequencies corresponding to the `gamma`-decays shown in figure.
Given, mass of `._12Mg^27` atom =26.991425 amu and mass of `._13Al^27` atom = 26.990080 amu

Explanation

Energy of photon `v_1`
`hv_1=(E_3-E_2)/h`
`=((1.015-0.834)MeV)/(6.62xx10^(-34) Js) `
`=(0.181xx1.6xx10^(-13) J)/(6.62xx10^(-34) Js) `
`=4.37xx10^(19) s^(-1)`
Energy of photon `v_2`
`v_2=(E_3-E_1)/h`
`=((1.015-0)MeV)/(6.62xx10^(-34) Js)`
`=2.45xx10^20 s^(-1)`
Energy of photon `v_3`
`v_3=(E_2-E_1)/h`
`=((0.834-0)MeV)/(6.62xx10^(-34))`
`=2.0xx10^20 s^(-1)`
Now emission of `beta_1^-` -particle is given by
`._12Mg^27 to ._13Al^27+ beta^(-) + v_2+Q`
`Q=[m(._12Mg^27)-m(._13Al^27) -E(v_2)]`
`=[26.991425-26.990080] u -(E_3-E_1)MeV`
=0.001345 x 931 - 1.015 MeV
`therefore` K.E. of `beta_1^-` particle =0.237