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 particle is `4.002amu` , determine the total kinetic energy in the final state Hence , obratain the mass of the parent nucleus in amu (1 amu = 931.470 MeV//e^(2))`

Answer

A, B, D

Explanation

Let the reaction be
`_(Z)^(A) rarr _(Z)^(A) _(-2)^(-4) Y + _(2)^(4) He`
Here `m_(y) = 223.61 amu and m_(a) = 4.002 amu`
we know that
`lambda(h)/(m v) rArr m^(2) v^(2) = (h^(2))/(lambda^(2)) = P^(2)`
: ` rArr But K.E. = (p^(2))/(2m) Therefore K.E. = (h^(2))/(2m lambda^(2)) ` .....(i)
Appling eq (i) for Y and a we get
`K.E_(a)= ((6.6 xx 10^(-34))^(2))/(2 xx 4.002 xx 1.67 xx 10^(-27) xx 5.76 xx 10^(-15) xx 5.76 xx 10^(-15))`
`= 0.0982249 xx 10^(-11) = 0.982 xx 10^(-12)J`
Similarly `(K.E.)_(y) = 0.0178 xx 10^(-12)J `
`Total energy = 10^(-12)J `
` we know that `E = deltamc^(2)`
`:. delta m = (E)/(c^(2)) = (10^(-12))/((3 xx 10^(8)^(2)) kg `
`1.65 xx 10^(-27) kg = 1 amu`
:' 65 xx 10^(-12))/((3 xx 10^(8)^(2)) kg = (10^(-12)atom )/(1.67 xx 10^(-27) xx (3 xx 10^(8)^(2))^(2))`
`= (10^(-12)atom )/(1.67 xx 10^(-27) xx (10^(16)) = 0.00665 amu `
The mass of the parent X will be
` m_(x) = m_(y) + m_(a) + deltam `
`= 223.61 + 4.002 + 0.00665 = 227.62 amu`