Ncert Exercises & Solutions

13.13 The radionuclide ¹¹C decays according to

\({ }_{6}^{11} \mathrm{C} \rightarrow{ }_{5}^{11} \mathrm{~B}+e^{+}+\nu:~~~ T_{1 / 2}=20.3 \min\)

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values:

m (\({ }_{6}^{11} \mathrm{C}\)) = 11.011434 u and m (\({ }_{6}^{11} \mathrm{~B}\)) = 11.009305 u,

Calculate Q and compare it with the maximum energy of the positron emitted.

Hint: Q-value can be found using mass defect of the nucleus.
Step 1: Find the equation for Q-value.
The given values are:

m ({}_{6}^{11}C) = 11.011434 u a n d m ({}_{6}^{11}B) = 11.009305 u,

The given nuclear reaction is:

{}_{6}^{11}C \to {}_{5}^{11}B + e^{+} + v

The half-life of

{}_{6}^{11}C

nuclei, \(T_{1 / 2}=20.3 \min\)
The maximum energy possessed by the emitted positron=0.960 MeV

The change in the Q-value,

(∆ Q)

of the nuclear masses of the

{}_{6}^{11}C

:
\(\Delta Q=\left[m^{\prime}\left({ }_{6}^{11} C\right)-\left[m^{\prime}\left\{{ }_{5}^{11} B\right\}+m_{e}\right]\right] c^{2}\)

where

m_{e}

=Mass of an electron or positron=0.00058 u

c=speed of light and m'=Respective nuclear masses.

Step 2: Find the value of Q-value.
If atomic masses are used instead of nuclear masses, we have to add 6

m_{e}

in the case of

{}_{6}^{11}C

and 5

m_{e}

in the case of

{}_{5}^{11}B

Hence, equation (1) reduces to:

∆ Q = [m ({}_{6}^{11}C) - m ({}_{5}^{11}B) - 2 m_{e}] c^{2}

Here,

m ({}_{6}^{11}C) a n d m ({}_{5}^{11}B)

are the atomic masses.

∴ ∆ Q = [11.011434 - 11.009305 - 2 \times 0.00054] c^{2} = (0.001033 c^{2}) u

But 1 u=931.5

M e V / c^{2}

∴ ∆ Q = 0.0010033 \times 931.5 \approx 0.962 M e V

The value of Q is almost comparable to the maximum energy of the emitted positron.