:

   

= 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(

   

) – m_e

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

Δm = m (

   

) – 6m_e – (m(

   

) – 5m_e )- m_e

= m(

   

) – 6m_e – m(

   

) + 5m_e – m_e

= m (

   

) – m(

   

) – 2 m_e

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.