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)`)

Answer

`6.25 MeV, 227.62 "amu"`

Explanation

(i) Given mass of `alpha`-particle, `m = 4.002` amu and mass of daughter nucleus `M = 223.610` amu, de-Broglie wavelength of `alpha`-particle,
`lambda = 5.76 xx 10^(-15) m`
So, momentum of `alpha`-particle would be
`p = (h)/(lambda) = (6.63 xx 10^(-34))/(5.76 xx 10^(15)) kg- m//s`
`rArr = 1.151 xx 10^(-19) kg - m//s`
From law of conservation of linear momentum, this should also be equal to the linear momentum the daughter nucleus (in opposite direction).
Let `K_(1)` and `K_(2)` be the kinetic energies of `alpha`-particle and daughter nucleus. Then total kinetic energy in the final state is `K = K_(1) + K_(2) = (p^(2))/(2m) + (p^(2))/(2M)`
`= (p^(2))/(2) ((1)/(m) + (1)/(M)) = (o^(2))/(2) ((M + m)/(Mm))`
1 amu `= 1.67 xx 10^(-27) kg`
Substituting the values, we get
`K = ((1.151 xx 10^(-19))^(2))/(2) xx ((M + m)/(M xx m))`
`= (p^(2))/(2) xx ((4.002 + 223.61) (1.67 xx 10^(-27)))/((4.002 xx 1.67 xx 10^(-27)) (223.61 xx 1.67 xx 10^(-27)))`
`K = 10^(-12) J = (10^(-12))/(1.6 xx 10^(-13)) MeV = 6.25 MeV`
`rArr K = 6.25 MeV`
(ii) Mass defect, `Deltam = (6.25)/(931.470)` amu `= 0.0067` amu
Therefore, mass of parent nucleus
`=` mass of `alpha`-particle `+` mass of daughter
nucleus `+` mass defect `(Delta m)`
`= (4.002 + 223.610 + 0.0067)` amu `= 227.62` amu