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Question

11C undergoes β+ decay to 11B. What is the maximum energy a positron emitted in the process can possess?
Atomic masses: 11C:11.0114 u; 11B:11.0093 u; e: 0.005486. c2=931MeVu.

A
Can't be uniquely determined.
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B
1.44 MeV
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C
833.6 keV
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D
933.6 keV
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Solution

The correct option is D 933.6 keV
Let us consider a general β+ decay process, as you have learned:
AZX AZ1Y+e++v.
Now, mass of the nucleus, AZX, is equal to the mass of the atom - mass of the electrons
=m(AZX)Zme.
Here m(AZX) is the atomic mass of AZX, Z is the atomic number (which is equal to the number of electrons) and me is the mass of an electron.
Similarly, mass of nucleus AZ1Y=m( AZ1Y)(Z1)me.
Now, mass of positron (e+) = mass of electron = me.
And, the neutrino, ν, is like a photon in that it has zero rest mass!
So, the total mass of the products = mass of nucleus
AZ1Y + mass of e+
=m( AZ1Y)(Z1)me+me
=m( AZ1Y)(Z+2)me.
Therefore, the mass lost = mass of parent nuclei - total mass of products
=m(AZX)Zmem( AZ1Y)(Z+2)me
=m(AZX)m( AZ1Y)2me.
And the equivalent energy associated with this lost mass, or the Q-Value
(=m(AZX)m( AZ1Y)2me)c2.
Here the parent and daughter nuclei are 11C and 11B
respectively. So, the QValue=(11.0114 u11.0093 u2×0.005486 u)×931MeVu.
=0.933607 MeV=933.6 keV.
Now, this energy is shared by the neutrino and positron, in general, in some ratio. Therefore, the maximum possible energy the positron can have is the total energy! This can happen when none of the energy is taken away by the neutrino. Hence, here, the maximum energy a positron can possess is 933.6 keV.

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