Question

A nucleus with mass number 229 emits a 3.443 MeV alpha particle. Calculate the disintegration energy Q for this process, taking the recoil energy of the residual nucleus into account.

Answer 1

3.504 MeV

Step-by-step explanation:

Given that;
`(A)/(Z) X ---->(A-4)/(Z-2) Y +\alpha + Q`

Also;

`Q= KE_(\alpha ) (M_(Y) + M_(\alpha ) /M_{Y`)

Mass number of X = 229

Mass number of Y = 225

Mass number of alpha particles = 4

Kinetic energy of alpha particles = 3.443 MeV

Q = 3.443 MeV (225 + 4/225)

Q= 3.504 MeV