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The radionuclide 11C decays according to

   

:

   

= 20.3
min

The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values:
m (116 C) = 11.011434 u and

m (116B ) = 11.009305 u,             
calculate Q and compare it with the maximum energy of the positron emitted.

Ans:

   

:

   

=
20.3 min

We know that the Mass defect in the reaction is given by

Δm = m (

   

) – m(

   

) – me

This is given in terms of atomic masses. We will express it in terms of nuclear mass, so we will subtract 6me  and 5me from carbon and boron respectively.

Δm = m (

   

) – 6me – (m(

   

) – 5me )- me

= m(

   

) – 6me – m(

   

) + 5me – me

= m (

   

) – m(

   

) – 2 me

Δm = [11.011434 – 11.009305 – 2 x 0.000548] u

= 0.002129 – 0.001096 = 0.001033

Q = Δm x 931 MeV

= 0.001033 x 931

Q= 0.9617 MeV

The Q-factor of the reaction is equal to the maximum energy of the emitted positron.