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# The radio nuclide 11C decays according to116C→115B+e++ν: T1/2=20.3 minThe maximum energy of the emitted positron is 0.960 MeV. Given the mass values:m(116C)=11.011434u and m(116B)=11.009305uCalculate Q and compare it with the maximum energy of the positron emitted.

Solution
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#### 116C→116B+e++ν+QQ=[mN(116C)−mN(116B)−me]c2Where, the masses used are those of nuclei and not of atoms. If we use atomic masses, we have to add 6me in case of 11C and 5me in case of 11B. Hence,Q=[m(116C)−m(116B)−2me]c2 (Note me has been doubled)Using given masses, Q = 0.961 MeV.Q = Ed+Ee+EνThe daughter nucleus is too heavy compared to e+ and ν, so it carries negligible energy (Ed ≈ 0). If the kinetic energy (Eν) carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q, hence, maximum Ee ≈ Q.

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Similar Questions
Q1
The radio nuclide 11C decays according to
116C115B+e++ν: T1/2 =20.3 min
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values:
m(116C)=11.011434u and m(116B)=11.009305u
Calculate Q and compare it with the maximum energy of the positron emitted.
View Solution
Q2
The radionuclide 11C decays according to11 11 +6 5 1/2 C B+ + : =20.3 min → e T νThe maximum energy of the emitted positron is 0.960 MeV.Given the mass values:m (116C) = 11.011434 u and m (116B ) = 11.009305 u,calculate Q and compare it with the maximum energy of the positron emitted.
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Q3

The radionuclide 11C decays according to

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values:

calculate Q and compare it with the maximum energy of the positron emitted

View Solution
Q4
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.
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Q5
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.
View Solution