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)`
`(0.181xx1.6 xx10^(-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) =2.45 xx10^(20)s^(-1)`
Energy of photon `v_3,v_3=(E_3-E_1)/h=((0.834-0)MeV)/(6.62xx10^(-34))=2.0xx10^(20)s^(-1)`
Now emission of `beta_1^-` particle is given by `""_(12)Mg^(27) to ""_(13)AI^(27) +beta^(-)+v_2+Q`
`Q=[m(""_(12)Mg^(27))-m(""_(13)AI^(27))-E(v_2)`
`=[26.991425-26.990080]u-(E_3-E_1)MeV`
`=0.001345xx931 -1.015 MeV`
`therefore` K.E of `beta_2^-` particle is given by `""_(12)Mg^(27) to ""_(13)AI^(27)+beta_2^(-)+v_3+Q`
`={[26.991425-26.990080]931-0.34}MeV`
=0.418 MeV