Consider the position emission reaction
`._(z)^(A)Xrarrunderset(z-1)AY+_(+)._(1)^(0)e+v`
Atomic mass of X and Y are `m_(x)` and `m_(y)` respectively and mass of a position is `m_(e)`
(a) Write the disintegration energy (Q) of the reaction.
(b) If `._(z)^(A)X` is `._(6)^(11)C` and atomic mass of `""^(11)C` is 11.011434 u, atomic mass of `""^(11)B` is 11.009305 u, mass of positron is `m_(e)=0.000549u`, then find the maximum kinetic energy of emitted positron. `[(1u)c^(2)=930MeV]`

5 views

1 Answers

Correct Answer - (a) `Q=[m_(x)-m_(y)-2m_(e)]C^(2)` (b) 0.959MeV

5 views

Related Questions