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

Answer:

3.504 MeV

Explanation:

Given that;$\frac{A}{Z} X ---->\frac{A-4}{Z-2} Y +\alpha + Q$

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