Q: Write symbolically the nuclear \beta^+ decay process of _{6}^{11}C . Is the decayed product X an isotope or isobar of ( _{6}^{11}C )? Given the mass values m ( _{6}^{11}C ) = 11.011434 u and m (X) = 11.009305 u. Estimate the Q-value in this process.

\beta^+ decay: _{116}C \rightarrow _{115} X+e^+ + \nu e^+= Positron \nu= NeutrinoAs mass number of C and X are the same, C and X are isobars.Energy released Q=(m_c-m_x-m_e)c^2 =[(m_c+116m_e)-(m_x+115m_e)]c^2 =[m_c-m_x]c^2 =[11.011434-11.009305]c^2 We have c^2=931.5 MeV/amu =0.002129 \times 931.5 MeV =1.983 MeV

Question 1

Define mass defect of a nucleus. Binding energy of  { _{ 8 }^{  }{ O } }^{ 16 }  is  127.5MeV . Write the value of its 'binding energy per nucleon'. Write the value of  1eV  energy in joule.

Accepted Answer

Mass defect is the amount by which the mass of an atomic nucleus differs from the sum of the masses of its constituent particles, being the mass equivalent of the energy released in the formation of the nucleus.Nuclear binding energy per nucleon for the given atom= \dfrac{BE}{A} =\dfrac{127.5MeV}{16} =7.96MeV Value of  1eV  is equal to  1.6	imes 10^{-19}J .

Question 2

State the law of radioactive decay. Hence derive the expression  N={ N }_{ 0 }{ e }^{ -\lambda t }  where symbols have their usual meanings.

Accepted Answer

Law of radioactive decay : At any instant, the rate of radioactive disintegration is directly proportional to the number of nuclei of the radioactive element present at that instant.We know by radioactive decay law \dfrac { dN }{ dt } \propto N or  \dfrac { dN }{ dt } =-\lambda N where  \lambda  is the constant of proportionality called radioactive decay constant. Let   { N }_{ 0 }  be the number of nuclei present at time  t=0 , and  N  the number of nuclei present at time  t . \int _{ { N }_{ 0 } }^{ N }{ \dfrac { dN }{ N }  } =-\int _{ 0 }^{ t }{ \lambda dt } =-\lambda \int _{ 0 }^{ t }{ dt }  \log _{ e }{ N } -\log _{ e }{ { N }_{ 0 } } =-\lambda t \log _{ e }{ \left( \dfrac { N }{ { N }_{ 0 } }  \right)  } =-\lambda t \dfrac { N }{ { N }_{ 0 } } ={ e }^{ -\lambda t } N={ N }_{ 0 }{ e }^{ -\lambda t }

Question 3

(a) Write the characteristic properties of nuclear force.
(b) Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from the graph

Accepted Answer

a)  The characteristic properties of nuclear force -i) The nuclear force is short-ranged strong attractive forces.ii) The nuclear forces between proton-proton, proton-neutron and neutron-neutron are approximately same.iii) The nuclear force is independent of electric charge.  b)  The potential energy (PE) vs separation graph is shown in the figure. The conclusions drawn from the graph are -i) The nuclear force is a short-range force.ii) The nuclear force is repulsive when the separation is less 1 fm  (1 fermi =10^{-15} m)  and it is attractive when the separation is greater than 1 fm.

Question 4

Bombardment of Beryllium by alpha particles resulted in the discovery of :

Accepted Answer

Bombardment of Beryllium by alpha particles reaction is given as below: _4^9Be + _2^4He \rightarrow [_6^{13}C] \rightarrow _6^{12}C + _0^1n So, correct choice is option C.

Question 5

(a) What do you mean by  Q  value of a nuclear reaction?(b) Write down the expression for  Q  value in the class of  \alpha  decay.

Accepted Answer

a) The Q-value of the reaction is defined as the difference between the sum of the masses of the initial reactants and the sum of the masses of the final products, in energy units (usually in MeV). It is the measure of amount of energy released by that reaction.b) The Q-value of the decay,  Q_{\alpha}  is the difference of the mass of the parent ( m_P ) and the combined mass of the daughter (( m_P ) and the  \alpha -particle ( m_{\alpha} ), multiplied by  c^2 : Q_{\alpha} = (m_P - m_D - m_{\alpha})c^2 .

Question 6

Plot a graph showing the variation of undecayed nuclei  N  versus time  t . From the graph, find out how one can determine the half-life and average life of the radioactive nuclei.

Accepted Answer

Law of Radioactivity defines that the number of Nuclei undergoing number of Nuclei present in the sample at that Instant.Since from the graph; we have      N=N_{ - }e^{-\lambda t }     ....1where  \lambda  =Disintegration constantFor  	herefore   For  T_{1/2}  is the time at  N-\frac{1}{2}N_D \Rightarrow   t=T_{1/2} \frac{N_0}{2} = N_0e^{-\lambda\frac{T}{2}} \Rightarrow   \boxed { T_{ 1/2 }=\frac { Ln2 }{ \lambda  } =\frac { 0.693 }{ \lambda   }  }  And for Mean life -we have to sum it over the whole Range for  N(t)=N_0e^{-\lambda t} for number of nuclei which decay in time t to t  t  	riangle ; N(t)	riangle t=\lambda N_{ 0 }e^{ -\lambda  \,  t}	riangle t  For Integration it over the Range  T=0\, to\,\infty 	au = \frac { \lambda N_{ 0 }\iint _{ 0 }^{ \infty  }{ te^{ -\lambda t }dt }  }{ N_ 0 } =\lambda \int ^{ \infty  }_0{ t } e^{ \lambda t }dt \boxed { 	au =\frac { 1 }{ \lambda }  }  Mean-life

Question 7

Explain the principle and working of a nuclear reactor with the help of a labelled diagram.

Accepted Answer

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Question 8

Distinguish between nuclear fission and fusion. Show how in both these processes energy is released. Calculate the energy release in MeV in the deuterium-tritium fusion reaction :

_{1}^{2} H +_{1}^{3} H \to_{1}^{4} H e +_{0} n^{1}

Using the data:

m (_{1}^{2} H) = 2.014102 u

m (_{1}^{3} H) = 3.016049 u

m (_{2}^{4} H e) = 4.002603 u

m n = 1.008665 u

l u = 931.5 M e V / c 2

Accepted Answer

Nuclear fission is a process of splitting of a heavier nucleus into two lighter nuclei. It generally occurs in elements of high atomic mass.Nuclear fusion is a process of combination of two light nuclei to form a heavier nuclei. It generally occurs in elements of low atomic mass.In both the processes, the total mass of the products is slightly less than the mass of the original nuclei. This difference in mass is converted to energy.In the given problem, mass defect is \Delta m  = 2.014102+3.016049-4.002603-1.008665 \Delta m = 0.018883\ u Energy released,  \Delta E = \Delta m c^2 \Delta E = 0.018883 	imes 931.5 = 17.589 MeV

Question 9

Why nuclear fusion reaction is also called thermo-nuclear reaction?

Accepted Answer

Thermonuclear fusion is a way to achieve nuclear fusion by using extremely high temperatures. So basically we can say nuclear fusion at high temperature will be thermo-nuclear reaction. Temperature is a measure of the average kinetic energy of particles, so by heating the material it will gain energy. After reaching sufficient temperature, given by the Lawson criterion, the energy of accidental collisions within the plasma is high enough to overcome the Coulomb barrier and the particles may fuse together.

Question 10

Write symbolically the nuclear  \beta^+  decay process of  _{6}^{11}C . Is the decayed product X an isotope or isobar of ( _{6}^{11}C )? Given the mass values m ( _{6}^{11}C ) = 11.011434 u and m (X) = 11.009305 u. Estimate the Q-value in this process.

Accepted Answer

\beta^+  decay: _{116}C  ightarrow _{115} X+e^+ + 
u e^+=  Positron 
u=  NeutrinoAs mass number of  C  and  X  are the same,  C  and  X  are isobars.Energy released  Q=(m_c-m_x-m_e)c^2 =[(m_c+116m_e)-(m_x+115m_e)]c^2 =[m_c-m_x]c^2 =[11.011434-11.009305]c^2 We have  c^2=931.5 MeV/amu =0.002129 	imes 931.5 MeV =1.983    MeV