A nucleus at rest undergoes a decay emitting an a particle of de - Broglie wavelength ` lambda = 5.76 xx 10^(-15)m ` if the mass of the daughter nucleus is 223.610 amu and that of alpha particle is `4.002amu` , determine the total kinetic energy in the final state Hence , obtain the mass of the parent nucleus in amu (1 amu = 931.470 `MeV//e^(2)`)

Explanation

`lambda_(d) = 5.76xx10^(-15)m`
`lambda_(d) = (h)/(p)rArrp = (h)/(lambda_(d)) = (6.6xx10^(-34))/(5.76xx10^(-15))`
`= 1.15xx10^(-19)kgm//sec`
`X rarrY + alpha`
By momentum conservation
`0 = p_(Y)+p_(alpha)=p_(Y)+p = 0`
`p_(Y)=- p`
Total `K.E.` in final state
`= (p^(2))/(2m_(Y))+(p^(2))/(2m_(alpha)) = (p^(2))/(2)[(m_(alpha)+m_(Y))/(m_(alpha)m_(Y))]`
`=(1.15xx10^(-19))^(2)/(2)[(4.002+223.610)/(4.002xx223.610)]xx(1)/(1.66xx10^(-27))`
`= 10^(-12)J = (10^(-12))/(1.6xx10^(-13)) = 6.25MeV`
`= (6.25)/(931)"amu" = 0.007"amu"`
Mass of parent nucleus
`m_(X) = m_(Y)+m_(alpha) = 0.007`
`=233.610 +4.002+0.007`
`= 227.619amu`