Q: The radio nuclide ^{11}C decays according to ^{11}_{6}C\, \rightarrow\, ^{11}_5B\, +\, e^+\, +\, \nu : T_{1/2} = 20.3\ min The maximum energy of the emitted positron is 0.960\ MeV . Given the mass values: m\, (^{11}_{6}C)\, =\, 11.011434\, u and m\, (^{11}_{6}B)\, =\, 11.009305\, u Calculate Q and compare it with the maximum energy of the positron emitted.

^{11}_{6}C\, \rightarrow\, ^{11}_{6}B\, +\, e^+\, +\, \nu\, +\, Q Q\, =\, \left [ m_N\left ( ^{11}_6C \right )\, -\, m_N\left ( ^{11}_6B \right )\, -\, m_e \right ]\, c^2 Where, the masses used are those of nuclei and not of atoms. If we use atomic masses, we have to add 6m_e in case of ^{11}C and 5m_e in case of ^{11}B . Hence, Q\, =\, \left [ m\left ( ^{11}_6C \right )\, -\, m\left ( ^{11}_6B \right )\, -\, 2m_e \right ]\, c^2 (Note m_e has been doubled)Using given masses, Q = 0.961 MeV.Q = E_d\, +\, E_e\, +\, E_\nu The daughter nucleus is too heavy compared to e^+ and \nu , so it carries negligible energy ( E_d \approx 0). If the kinetic energy ( E_\nu ) carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q, hence, maximum E^e \approx Q.

Q: In a radioactive sample ^{40}_{19}K nuclei either decay into stable {^{40}_{20}Ca} nuclei with decay constant 4.5\times 10^{-10} per year or into stable ^{40}_{18}Ar nuclei with decay constant 0.5\times 10^{-10} per year. Given that in this sample all the stable ^{40}_{20}Ca and ^{40}_{18}Ar nuclei are produced by the ^{40}_{19}K nuclei only. In time t\times 10^{9} years, if the ratio of the sum of stable ^{40}_{20}Ca and ^{40}_{18}Ar nuclei to the radioactive ^{40}_{19}K nuclei is 99 , the value of t will be? [Given: In 10=2.3 ]

t=0 \dfrac{dN}{dt}=2\left(\lambda_1+\lambda_2\right)\times N log_e\left(\dfrac{N}{N_0}\right)=-(\lambda_1+\lambda_2)t 2.3\times log_{10}\left(\dfrac{N_0}{N_0/100}\right)=5\times 10^{-10}t \dfrac{2.303\times 2}{5\times 10^{-10}}=t 2.303\times 0.4\times 10^{10}=t t=9.2\times 10^9 year.

Question 1

A free nucleus of mass  24 amu  emits a gamma photon (when initially at rest). The energy of the photon is  7 MeV . The recoil energy of the nucleus in  keV  is

Accepted Answer

E = \cfrac{p^2}{2m} Conservation of momentum for photon: E = \cfrac{hc}{\lambda} = 7 MeV p = \cfrac{h}{\lambda} = 7/c MeV Equating the momentum:  \cfrac{7}{c }= \sqrt{2E_{nucleus}m} Substitute  m = 24\ amu ,Solving with appropriate units:  E_{nucleus} = 1.1\  KeV

Question 2

The radio nuclide  ^{11}C  decays according to ^{11}_{6}C\, ightarrow\, ^{11}_5B\, +\, e^+\, +\, 
u  :   T_{1/2}   = 20.3\ min The maximum energy of the emitted positron is  0.960\ MeV . Given the mass values: m\, (^{11}_{6}C)\, =\, 11.011434\, u  and  m\, (^{11}_{6}B)\, =\, 11.009305\, u Calculate  Q  and compare it with the maximum energy of the positron emitted.

Accepted Answer

^{11}_{6}C\, ightarrow\, ^{11}_{6}B\, +\, e^+\, +\, 
u\, +\, Q Q\, =\, \left [ m_N\left ( ^{11}_6C ight )\, -\, m_N\left ( ^{11}_6B  ight )\, -\, m_e ight ]\, c^2 Where, the masses used are those of nuclei and not of atoms. If we use atomic masses, we have to add  6m_e  in case of  ^{11}C  and  5m_e  in case of  ^{11}B . Hence, Q\, =\, \left [ m\left ( ^{11}_6C ight )\, -\, m\left ( ^{11}_6B  ight )\, -\, 2m_e ight ]\, c^2  (Note  m_e  has been doubled)Using given masses, Q = 0.961 MeV.Q =  E_d\, +\, E_e\, +\, E_
u The daughter nucleus is too heavy compared to  e^+  and  
u , so it carries negligible energy ( E_d   \approx  0). If the kinetic energy ( E_
u ) carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q, hence, maximum  E^e   \approx  Q.

Question 3

A nucleus with mass number  220  initially at rest emits an  \alpha \ particle . If the  Q  value of the reaction is  5.5 MeV , calculate the kinetic energy of the  \alpha\ particle

Accepted Answer

M \rightarrow X+ \alpha  E = \cfrac{p^2}{2m} Let Total energy =  T  T = E_{\alpha} + E_{X} M = M_{\alpha} + M_{X} M_{X}= 216 Conservation of momentum  p_{\alpha} = p_{X}  m_{\alpha} E_{alpha} = m_XE_X E_{\alpha} = \cfrac{m_x}{M}T  E_{\alpha} = \cfrac{216}{220} 5.5  = 5.4 MeV

Question 4

In a radioactive sample  ^{40}_{19}K  nuclei either decay into stable  {^{40}_{20}Ca}  nuclei with decay constant  4.5	imes 10^{-10}  per year or into stable  ^{40}_{18}Ar  nuclei with decay constant  0.5	imes 10^{-10}  per year. Given that in this sample all the stable  ^{40}_{20}Ca  and  ^{40}_{18}Ar  nuclei are produced by the  ^{40}_{19}K  nuclei only. In time  t	imes 10^{9}  years, if the ratio of the sum of stable  ^{40}_{20}Ca  and  ^{40}_{18}Ar  nuclei to the radioactive  ^{40}_{19}K  nuclei is  99 , the value of t will be? [Given: In  10=2.3 ]

Accepted Answer

t=0 \dfrac{dN}{dt}=2\left(\lambda_1+\lambda_2ight)	imes N log_e\left(\dfrac{N}{N_0}ight)=-(\lambda_1+\lambda_2)t 2.3	imes log_{10}\left(\dfrac{N_0}{N_0/100}ight)=5	imes 10^{-10}t \dfrac{2.303	imes 2}{5	imes 10^{-10}}=t 2.303	imes 0.4	imes 10^{10}=t t=9.2	imes 10^9  year.

Question 5

Suppose India had a target of producing by   2020 \mathrm{AD}, 200000  MW  of electric power,  10     \%    of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilisation (i.e.conversion to electric energy) of thermal energy produced in a reactor was  25     \%    How much amount of fissionable uranium would our country need per year by  2020    ?    Take  235    \mathrm{U}    to be about  200    \mathrm{MeV}    .

Accepted Answer

Total electric power which is aimed to be produced by 2020 =  2 	imes 10^5 \ MW Power obtained from the nuclear power plant =   10 	ext{% of } 2 	imes 10^5 \ MW = 2 	imes 10^4 \ MW= 2 	imes 10^{10} \ W  Energy required from the nuclear plant in one year=   	ext{Power} 	imes 	ext{time} = 2 	imes 10^{10} 	imes 365 	imes 24 	imes 60 	imes 60 = 6.3 	imes 10^{17} \ J Available electric energy per fission=  25 	ext{% of } 200 \ MeV = (0.25 	imes 200 	imes 1.602 	imes 10^{-13}) \ J = 8 	imes 10^{-12} \ J Required no. of fission per year =  \dfrac{6.3 	imes 10^{17} }{8 	imes 10^{-12}} = 0.7875 	imes 10^{29} Now,  6.023 	imes 10^{23}   nuclei of  _{92}^{235} U   have mass =  235 g 	herefore   Mass required to produce   7.9 	imes 10^{28}   nuclei =  \dfrac{ 235}{ 6.023 	imes 10^{23} } 	imes  3.95 	imes 10^{28} g \approx 3.084 	imes 10^4 \ kg

Question 6

A radioactive sample can decay by two different processes. The half-life for the first process is

T_{1}

and that for the second process is

T_{2}

. The effective half-life

T

of the radioactive sample is

Accepted Answer

Radioactive decay rate is directly proportional to number of nucleus present at any time \dfrac{dN}{dt} = -{\lambda}N  ....(i)where  \lambda  is a constant related to half life as  \lambda =\dfrac{\ln 2}{T_\frac{1}{2}}  where  T_\frac{1}{2}  is half life time of radioactive decayThe total rate of decay is \dfrac{dN}{dt} =- {(\lambda_1 +\lambda_2)}N =\lambda_{effective}N ...(ii) \lambda_1 =\dfrac{\ln 2}{T_1} \lambda_2 =\dfrac{\ln 2}{T_2} \lambda_{effective} =\dfrac{\ln 2}{T_{effective}} substituting in equation (ii) (\dfrac{\ln 2}{T_1}   +\dfrac{\ln 2}{T_2})N   =\dfrac{\ln 2}{T_{effective}}N \dfrac{1}{T}  =\dfrac{1}{T_1} +\dfrac{1}{T_2}

Question 7

Consider the D-T reaction (deuterium-tritium fusion)

_{1}^{2} C e +_{1}^{3} R u \to_{2}^{4} H e + n

Consider the radius of both deuterium and tritium to be approximately

2.0 f m

. What is the kinetic energy need to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion even = average thermal kinetic energy available with the interacting particles

2 (3 k T / 2)

;

k =

Boltzman's constant;

T =

absolute temperature

Accepted Answer

The potential energy of D-T system, when the two nuceli almost comes in contact with other is given by P.E=\cfrac { 1 }{ 4\pi { \in  }_{ 0 } } 	imes \cfrac { { q }_{ 1 }{ q }_{ 2 } }{ r }  { q }_{ 1 }=1.6	imes { 10 }^{ -19 }C  (Charge on deuterium nucleus) { q }_{ 2 }=1.6	imes { 10 }^{ -19 }C  (Charge on tritium nucleus) r=2+2=4fm=4	imes { 10 }^{ -15 }m  (Sum of radii of two nuclei) \quad P.E=9	imes { 10 }^{ 9 }	imes \cfrac { { \left( 1.6	imes { 10 }^{ -19 } ight)  }^{ 2 } }{ 4	imes { 10 }^{ -15 } } =5.764	imes { 10 }^{ -14 }J The kinetic energy required to overcome coulomb's repulsion is equal to the potential energyThe required kinetic energy  5.76	imes { 10 }^{ -14 }J At temperature  T , the system will possess kinetic energy equal to  \cfrac { 3 }{ 2 } kT , where  k  is Boltzmann's constantRequired temperature is given by \cfrac { 3 }{ 2 } kT=5.76	imes { 10 }^{ -14 }J\Rightarrow T=\cfrac { 2 }{ 3 } 	imes \cfrac { 5.76	imes { 10 }^{ -14 } }{ k } =\cfrac { 2	imes 5.76	imes { 10 }^{ -14 } }{ 3	imes 1.38	imes { 10 }^{ -23 } } =2.78	imes { 10 }^{ 9 }K

Nuclei