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

de Brogile wavelength of `alpha`-particles, `lambda_alpha=5.76 xx 10^(-15) m`
de Broglie wavelength, `lambda=(h)/(p)`
Therefore, momentum of `alpha`-particle,
`P_(alpha)=(h)/(lambda) (=6.63 xx 10^(-34))/(5.76xx10^(-15))1.15 xx 10^(-19) kg s^(-1)`
If `P_(d)` is momentum of daughter nucleus, then from coservation of linear momnetum,
`P_(d)= -p_(alpha) =- 1.15 xx10^(-19) kg m s^(-1)`
Hence, total kinetic of final state,
`E=E_(alpha) +E_(d) =(p_(alpha)^(2))/(2m_(alpha))+(p_(alpha)^(2))/(2m_(alpha))+(p_(d)^(2))/(2m_(d))`
`=(P_(alpha)^(2))/(2)((1)/(m_(alpha))+(1)/(m_(d))) =(P_(alpha)^(2)(m_(d)+m_(alpha)))/(2m_(alpha)md)`
`1 am u =931.470 MeV//c^(2)=(931.470 xx1.6 xx 10^(-13))/(3 xx 10^(8))^(2) kg`
`=1.66 xx 10^(-27) kg`
Therefore, total kinetic energy of final state is
` ((1.15xx10-^(19))^(2)xx(223.610+4.002) am u)/(2xx (4.00 am u) xx (223. 610 am u))`
`((1.15xx10^(-19))2xx227.612)/(2xx4.002 xx 223. 610 xx 1. 66 xx 10^(27))J`
`((1.15)^(2)xx227.612)/(2xx4.002 xx 223.610xx1.66) xx10^(11) J`
`=1^(-12) J=(10^(-12))/(1.6xx10^(-13)) MeV=6.25 MeV`
Mass of parent nucleus `= m_(d) +m_(alpha)-(BE)`
`=(223.610+4.002-(6.25)/(931.47))am u`
=(227. 612 -0.007) am u =227 .605 am u`.