Q: In \beta^- decay, a

The reaction of beta decay is shown as : n \rightarrow p + e^- + \bar{\nu}_e (antineutrino)So, in a \beta- decay, a neutron converts into a proton emitting antineutrino.option (A) is correct.

Question 1

In  \beta^-  decay, a

Accepted Answer

The reaction of beta decay is shown as : n ightarrow p + e^- + \bar{
u}_e                (antineutrino)So, in a  \beta-  decay, a neutron converts into a proton emitting antineutrino.option (A) is correct.

Question 2

When two nuclei  (A \le 10)  fuse together to form a heavier nucleus, the

Accepted Answer

Since the two nuclei fuse together to from a stable heavier nucleus, the heavier nucleus will be more stable that the two nuclei. So, the binding energy of the heavier nucleus is greater than the binding energy of the two nuclei.Option (A) is correct.

Question 3

For a radioactive substance, show the variation of the total mass disintegrated as a function of time t graphically.

Accepted Answer

For a radioactive substance total mass disintegrated as a function of time (t) is given as  M = M_0 e^{-\lambda \ t} Where,  M_0 =   mass of radioactive substance at  t= 0 \sec M =   Mass of radioactive substance present at any time  't''

Question 4

Initial mass of a radioactive substance is 3.2 mg. It has a half-life of 4 h. Find the mass of the substance left undecayed after 8 h.

Accepted Answer

Given data :-Initial Mass  = M_0 = 3.2 mg t_{1/2} = 4 m To calculate , mass of the substance left undecayed after  8h \, [M_z = 8] we know, M_{(z)} = M_0 e^{-\lambda t} \dfrac{M_0}{2} = M_0 e^{-\lambda (4)} \Rightarrow \lambda = \dfrac{1}{4} \ln (2) M_{(8)} = M_0 . e^{-\lambda . 8} \Rightarrow M_{(8)} = M_0 . e^{-\dfrac{1}{4} . \ln (2) . 8} M_{(8)} = M_0 e^{-2 \ln (2)} M = \left(\dfrac{M_0}{4} \right) = \dfrac{3.2 mg}{4} M_{(8)} = 0.8 mg

Question 5

Differentiate between half life and average life of a radioactive substance.

Accepted Answer

The difference between half life and average life is given below :- Half lifeAverage life (i) The time required for half of the radioactive substance to decay is called as half life of that substance.(ii) It is denoted by  t _{\frac{1}{2}} (iii)  t_{\frac{1}{2}} =   half life  = \left(\dfrac{\ln 2}{\lambda} ight) where  \lambda  is decay constant.(i) The expected life for the radioactive substance within a sample is called as average life of that substance.(ii) It is denoted by  	au .(iii)  	au =  average life  = \dfrac{1}{\lambda}

Question 6

A radioactive substance decays for an interval of time equal to its mean life. Find the fraction of the amount of the substance which is left undecayed after this time interval.

Accepted Answer

Let decay constant of Radio active Substance  = \lambda t_{mem} = t_{avg} = \dfrac{1}{\lambda} Let No is initial Amount  80  Amount remains After time t N_t = N_0 e^{-\lambda t} \dfrac{N_t}{N_0} = e^{-\lambda t}  ___(I) t = t_{avg} = \dfrac{1}{\lambda} from equation (I) \dfrac{N_t}{N_0} = e^{-\lambda 	imes \frac{1}{\lambda}} = e^{-1} \dfrac{N_t}{N_0} = \dfrac{1}{e} = 0.367 	herefore   fraction of Amount left \dfrac{N_t}{N_0} = 0.367

Question 7

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 8

Write two distinguishing feature of nuclear forces.

Accepted Answer

Two Distinguishing features of Nucleas forces :-(i) Nuclear focus are short Range focus. They are appreciable at a distance of  10^{-15}m  is less . At greater distance than this; they are negligiable (ii) Nuclear forces are independent of electric charges. Hence; the Nuclear Interaction between two protons ,  two neutrons and one one proton, one neutron is all same.

Question 9

Complete the following nuclear reaction for  \alpha  and  \beta  decay:(i)  _{92}^{238} U \longrightarrow\, ? + \, ^4_2He + Q (ii)  _{11}^{22} Na \longrightarrow\, _{10}^{22}Ne + ? + v

Accepted Answer

i)   _{ 92 }^{ 238 }{ U }\longrightarrow ?+_{ 2 }^{ 4 }{ He }+\lambda  Since  'U'  gines Helium  +'Q' 	herefore   It gives:- _{ 92 }^{ 238 }{ U }\longrightarrow _{ 90 }^{ 234 }{ Th }+_{ 2 }^{ 4 }{ He }+Q  ii)   _{ 11 }^{ 22 }{ Na }\longrightarrow _{ 10 }^{ 22 }{ Ne }+?+V \beta  Particle will be given. _{ 11 }^{ 22 }{ Na }\longrightarrow _{ 11 }^{ 22 }{ Ne }+\beta^++V

Question 10

(a) Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (BE/A) versus the mass number

A

.
(b) A radioactive isotope has a half-life of

10

years. How long will it take for the activity to reduce to

3.125

%?

Accepted Answer

A. The value of binding energy per nucleon gives a measure of the stability of that nucleus greater is the binding energy per nucleon of a nucleus , more stable is the nucleus.The above graph shown, binding energy per nucleon drawn against mass number A.The saturation effect of nuclear forces is the property responsible for the approximate constancy of binding energy in range 30<A<17030<A<170It is clear from the curve that binding energy per nucleon of the fused nucle is more than there of the light nuclei taking part nuclear fusion .Hence energy gets released in the process.In a nuclear fission , the sum of the masses of the final products is less than the sum of the masses of the reactant componentB.  N = N_{o} e ^{- \lambda t}    But   N = \dfrac{N_{o}}{32} \dfrac{N_{o}}{32} = N_{o} e^{-\lambda t} 2^{-5} = e^{-\lambda t} 5 ln2 = \lambda t but given half life so  \lambda = \dfrac{ln2}{10} so t = 50 Years

Question 11

Four nuclei of an element undergo fusion to form a heavier nucleus, with release of energy. Which of the two - the parent or the daughter nucleus - would have higher binding energy per nucleon?

Accepted Answer

Since the unstable parent nuclei fuse to form a heavier stable daughter nuclei in a nuclear fusion reaction releasing some energy. So, daughter nuclei is more stable than parent nuclei. Thus daughter nucleus has more binding energy per nucleon than parent nucleus.

Question 12

(a) Derive the mathematical expression for law of radioactive decay for a sample of a radioactive nucleus.(b) How is the mean life of a given radioactive nucleus related to the decay constant?

Accepted Answer

a) According to radioactive decay law, the rate of decay is proportional to amount of nuclei ,N. Thus,  -\dfrac{dN}{dt}\propto N   (negative sign come because the amount of nuclei decreases with time)So,  \dfrac{dN}{dt}=-\lambda N  or  \dfrac{dN}{N}=-\lambda t Integrating,  lnN=-\lambda t+c   where C is the integrating constant.At  t=0 ,  N=N_0  so  lnN_0=c Now,  lnN=-\lambda t+lnN_0 or  ln(N/N_0)=-\lambda t  or  N/N_0=e^{-\lambda t} Hence,  N=N_0e^{-\lambda t} , this is the mathematical expression of radioactive decay of nuclei. b) The relation between mean life  (t_{avg})  and decay constant  (\lambda)  :   t_{avg}=\dfrac{1}{\lambda}

Question 13

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 14

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