Carbon-11 decays to boron-11 according to the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The decay reaction is ${ }_{6}^{11} C \rightarrow{ }_{5}^{11} B +e^{+}+ v _{e}+0.96 \,MeV$.When the positron $e^{+}$ annihilates with an electron

Vedclass · Accepted Answer

$8 	imes 10^{16}$

Vedclass · Answer

(B) The decay reaction is ${ }_{6}^{11} C ightarrow{ }_{5}^{11} B +e^{+}+ v _{e}+0.96 \,MeV$.When the positron $e^{+}$ annihilates with an electron $e^{-}$,the energy released is $2 	imes m _{e} c^{2} = 2 	imes 0.511 \,MeV \approx 1.02 \,MeV$.Total energy released per decay = $0.96 \,MeV + 1.02 \,MeV = 1.98 \,MeV \approx 2 \,MeV$.Initial mass $M _{0} = 1 \,\mu g = 10^{-6} \,g$ of ${ }_{6}^{11} C$.Number of atoms $N _{0} = \frac{M _{0}}{A} 	imes N _{A} = \frac{10^{-6}}{11} 	imes 6.023 	imes 10^{23} \approx 5.475 	imes 10^{16} \,atoms$.In time $t = 2 t _{0}$,the fraction of atoms decayed is $1 - (1/2)^{2} = 1 - 1/4 = 3/4 = 0.75$.Number of atoms decayed $N = 0.75 	imes N _{0} = 0.75 	imes 5.475 	imes 10^{16} \approx 4.1 	imes 10^{16} \,atoms$.Total energy $E = N 	imes 1.98 \,MeV \approx 4.1 	imes 10^{16} 	imes 2 \,MeV \approx 8.2 	imes 10^{16} \,MeV$.Given the options,the closest value is $8 	imes 10^{16} \,MeV$.

Similar Questions

In a nuclear power station,uranium is used for the production of .......

In nuclear fission,the percentage of mass converted into energy is about: (in $\%$)

Control rods used in nuclear reactors are made of:

The source of the Sun's energy is

$A$ power of $100 \ W$ is produced by the fission of $1 \ kg$ of $^{235}U$. For approximately how long will the energy production continue?

Vedclass Test Series

Exam Paper Generator

Online Exam Module