Q: Consider the nuclear fission, Ne^{20} \rightarrow 2He^4 + C^{12} Given that the binding energy/nucleon of Ne^{20} , He^4 and C^{12} are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement:

Energy absorbed; \Delta Q = \text{Binding energy of products - Binding energy of reactants} \text{Binding energy of products} = 2 \times (\text{B.E of } He^4 ) + (\text{B.E of } C^{12} ) = 2 (4 \times 7.07 ) + (12 \times 7.86) =150.88 \ MeV \text{Binding energy of Reactants} = 20 \times 8.03 = 160.6 \ MeV \Delta Q = \text{Binding energy of products - Binding energy of reactants} = 150.88 - 160.6 = -9.72 \ MeV \text{ Hence, 9.72 MeV energy will be released}

Q: ^{131}I is an isotope of Iodine that \beta decays to an isotope of Xenon with a half-life of 8 days. A small amount of a serum labelled with ^{131}I is injected into the blood of a person. The activity of the amount of ^{131}I injected was 2.4\times 10^5 Becquerel (Bq). It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After 11.5 hours, 2.5 ml of blood is drawn from the person's body, and gives an activity of 115 Bq. The total volume of blood in the person's body, in liters is approximately (you may use e^x \approx 1 + x for |x| << 1 and ln 2 \approx 0.7 ).

t_\frac{1}{2} = 8 \ days = 8 \times 24 \ hr R_0 = 2.4 \times 10^5 Bq Using \ \dfrac{0.691}{t_\frac{1}{2}} \times t = ln\dfrac{R_0}{R} \dfrac{0.692}{8 \times 24} \times 11.5 =ln \dfrac{2.4 \times 10^5}{R} \dfrac{2.4 \times 10^5}{R} = e^{0.041} = 1+0.041 R = \dfrac{2.4 \times10^5}{1.041} = 2.3 \times 10^5 115 \ Bq \ is \ in\ volume = 2.5 ml 2.3 \times10^5 Bq \ is \ in\ volume =\dfrac{2.5}{115} \times 2.3 \times 10^5 = 0.05 \times 10^5 = 5 \times10^3 ml = 5\ litres

Question 1

The activity of a radioactive sample falls from  700 s^{-1}  to  500 s^{-1}  in  30  minutes. Its half life is close to :

Accepted Answer

From equation of radioactivity we know A_t = A_0 e^{-\lambda t} Where  A_t =   Amount of material at time 't' A_0 =  Amount of substance at  t = 0 \lambda =   decay constant t =   time.Taking log we have  \ln \left[\dfrac{A_0}{A_t} ight] = \lambda t For half life  A_t = \dfrac{A_0}{2} . \Rightarrow \ln \ 2 = \lambda t_{1/2}  __(A)  t_{1/2} =   half lifeFrom given condition. \ln \left[\dfrac{700}{500} ight] = \lambda (30 \ min)  __(B) (A) - (B) \dfrac{\ln 2}{\ln [7/5]} = \dfrac{t_{1/2}}{30} \Rightarrow t_{1/2} = \dfrac{\ln 2 	imes 30}{\ln [7/5]} \Rightarrow t_{1/2} = \dfrac{0.693 	imes 30}{0.336} \Rightarrow t_{1/2} = 61.8 min option (B) is correct.

Question 2

Half lives of two radioactive nuclei  A  and  B  are  10  minutes and  20  minutes, respectively. If, initially a sample has equal number of nuclei, then after  60  minutes, the ratio of decayed numbers of nuclei A and B will be:

Accepted Answer

N_A = N_{OA} e^{-\lambda t} = \dfrac{N_{OA}}{2^{t/t}\dfrac{1}{2}} = \dfrac{N_{OA}}{2^6} 	herefore  Number of nuclei  decayed = N_{OA} - \dfrac{N_{OA}}{2^6} = \dfrac{63N_{OA}}{64} N_B = N_{OB}e^{-\lambda t} = \dfrac{N_{OB}}{2^{t/t} \dfrac{1}{2}} = \dfrac{N_{OB}}{2^3} 	herefore  Number of nuclei decayed = N_{OB} - \dfrac{N_{OB}}{2^3} = \dfrac{7N_{OB}}{8} Since  N_{OA} = N_{OB} 	herefore  ratio of decayed numbers of nuclei A \& B = \dfrac{63 N_{OA} 	imes 8}{64 	imes 7N_{OB}} = \dfrac{9}{8}

Question 3

Two radioactive substances A and B have decay constants  5\lambda  and  \lambda  respectively. At  t=0  they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be  \left(\displaystyle\frac{1}{e}\right)^2  after a time interval

Accepted Answer

Decay equation is given by  N= {N}_{o}{e}^{-\lambda t} Where,  \lambda  is Decay constant;  N  is number of atom at time  t So, number of atom of  A  is given by,  {N}_{A}= {N}_{o}{e}^{-5\lambda t} And that of  B  are given by,  {N}_{B}= {N}_{o}{e}^{-\lambda t} Taking ration we get:  \dfrac{{N}_{A}}{{N}_{B}}={e}^{-4\lambda t} Given that:  \dfrac{{N}_{A}}{{N}_{B}}={e}^{-2} So, this will happen when,  t=\dfrac{1}{2\lambda}

Question 4

Two radioactive materials  A  and  B  have decay constants  10\lambda  and  \lambda , respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of  A  to that of  B  will be  1/e  after a time:

Accepted Answer

N_1=N_0e^{-10\lambda t} N_2=N_0e^{-\lambda t} \dfrac{1}{e}=\dfrac{N_1}{N_2}=e^{-9\lambda t} \Rightarrow 9\lambda t=1 t=\dfrac{1}{9\lambda}

Question 5

At a given instant, say t = 0, two radioactive substances A and B have equal activities. The ratio  \dfrac{R_B}{R_A}  of their activities after time t itself decays with time t as  e^{-3t}.  If the half-life of A is  ln2 , the half-life of B is:

Accepted Answer

Half life of  A = ln2 t_{1/2} = \dfrac{\ell n2}{\lambda} \lambda_A = 1 at  t = 0 \, \, R_A = R_B N_A e^{-\lambda AT} = N_B e^{-\lambda BT} N_A = N_B  at t = 0at t= t  \dfrac{R_B}{R_A} = \dfrac{N_0 e^{-\lambda_B^t}}{N_0 e^{-\lambda_A^t}} e^{-(\lambda_B -\lambda_A)t} = e^{-3t} \lambda_B - \lambda_A = 3 \lambda_B = 3 + \lambda_A = 4 t_{1/2} = \dfrac{\ell n2}{\lambda_B} = \dfrac{\ell n2}{4}

Question 6

Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At

t = 0

it was

1600

counts per second and

t = 8

seconds it was

100

counts per second. The count rate observed, as counts per second, at

t = 6

seconds is close to

Accepted Answer

At  t = 0, A_{0} = \dfrac {dN}{dt} = 1600\ C/s at  t = 8s, A = 100\ C/s \dfrac {A}{A_{0}} = \dfrac {1}{16}  in  8\ sec Therefore half life is  t_{1/2} = 2\ sec 	herefore  Activity at  t = 6  will be  1600\left (\dfrac {1}{2}ight )^{3} = 200\ C/s 	herefore  correct answer is  (3) .

Question 7

Consider the nuclear fission,  Ne^{20} \rightarrow 2He^4 + C^{12} Given that the binding energy/nucleon of  Ne^{20} ,  He^4  and  C^{12}  are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement:

Accepted Answer

Energy absorbed; \Delta Q = 	ext{Binding energy of products - Binding energy of reactants} 	ext{Binding energy of products} = 2 	imes (	ext{B.E of  }  He^4 ) + (	ext{B.E of  }  C^{12} ) = 2 (4 	imes 7.07 ) + (12 	imes 7.86) =150.88 \ MeV 	ext{Binding energy of Reactants}  =  20 	imes 8.03 = 160.6 \ MeV \Delta Q = 	ext{Binding energy of products - Binding energy of reactants} = 150.88 - 160.6 = -9.72 \ MeV 	ext{ Hence, 9.72 MeV energy will be released}

Question 8

At some instant, a radioactive sample  S_{1}  having an activity  5\ \mu Ci  has twice the number of nuclei as another sample  S_{2}  which has an activity of  10\ \mu Ci . The half lives of  S_{1}  and  S_{2}  are

Accepted Answer

For sample 1 we can write, N_1 = {N_o}_1e^{-\lambda_1 t} N_2 = {N_o}_2e^{-\lambda_2 t} Given that at a particular time t  N_1 = 2 N_2 Also given thatFor sample 1,  A_1 = -\dfrac{dN_1}{dt}= {N_o}_1 \lambda_1e^{-\lambda_1 t}= 5 \mu C_i A_2 = -\dfrac{dN_2}{dt}= {N_o}_2\lambda_2 e^{-\lambda_2t}= 10 \mu C_i Solvong above equations we get, \dfrac{\lambda_1}{\lambda_2}= \dfrac{1}{4} \dfrac{{half \ life}_1}{{half \ life}_2}=\dfrac{ln 2}{\lambda_1}	imes \dfrac{\lambda_2}{ln 2}= 4 T_{S_1}=4T_{S_2} \Rightarrow T_{S_1}=20  years and  T_{S_{2}}=5  years is the possible answer.

Question 9

A solution containing active cobalt  _{ 27 }^{ 60 }{ Co }  having activity of  0.8\mu Ci  and decay constant  \lambda  is injected in an animal's body. If  1{cm}^{3}  of blood is drawn from the animal's body after  10  hrs of injection, the activity found was  300  decays per minute. What is the volume of blood that is flowing in the body? ( 1Ci=3.7	imes {10}^{10}  decay per second and at  t=10hrs   { e }^{ -\lambda t }=0.84 )

Accepted Answer

The activity equation can be written as  -\dfrac{dN}{dt}= \lambda N_o e^{-\lambda t} given that \lambda N_o = 0.8 \mu C_i Putting the values, \lambda N_o = 2.96 	imes 10^{4}  Let the volume of the blood flowing be V,the activity would reduce by a factor of  \dfrac{10^{-3}}{V} Hence \dfrac{\lambda N_o 10^{-3}}{V}e^{-\lambda t}=300/60   (Both R.H.S. and L.H.S. are decay/s)Putting the values of  e^{-\lambda t}  and  \lambda N_o  we get V= 5 litre

Question 10

^{131}I  is an isotope of Iodine that  \beta  decays to an isotope of Xenon with a half-life of 8 days. A small amount of a serum labelled with  ^{131}I  is injected into the blood of a person. The activity of the amount of  ^{131}I  injected was  2.4	imes 10^5  Becquerel (Bq). It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After  11.5  hours,  2.5 ml  of blood is drawn from the person's body, and gives an activity of  115  Bq. The total volume of blood in the person's body, in liters is approximately (you may use  e^x \approx  1 + x  for  |x| << 1  and  ln 2 \approx  0.7 ).

Accepted Answer

t_\frac{1}{2} = 8 \ days = 8 	imes 24 \ hr  R_0 = 2.4 	imes 10^5 Bq  Using  \ \dfrac{0.691}{t_\frac{1}{2}} 	imes t = ln\dfrac{R_0}{R}  \dfrac{0.692}{8 	imes 24} 	imes 11.5 =ln \dfrac{2.4 	imes 10^5}{R}   \dfrac{2.4 	imes 10^5}{R} = e^{0.041} = 1+0.041    R = \dfrac{2.4 	imes10^5}{1.041} = 2.3 	imes 10^5  115 \ Bq \  is \ in\  volume = 2.5 ml 2.3 	imes10^5 Bq \ is \ in\  volume =\dfrac{2.5}{115} 	imes 2.3 	imes 10^5    =  0.05 	imes 10^5 = 5 	imes10^3 ml  = 5\ litres

Question 11

In an experiment the initial number of radioactive nuclei is  3000 . It is found that  1000\pm 40  nuclei decayed in the  1.0\ s . For  |x| < < 1, \ln(1+x) = x  up to first power in  x . The error  \Delta \lambda , in the determination of the decay constant  \lambda , is  s^{-1} , is

Accepted Answer

N = N_{o} e ^{-\lambda t} &#10;  ln N = lnN_{o} - \lambda t &#10; \dfrac{dN}{N} = - \lambda t &#10;Converting to error&#10; \dfrac{\Delta N}{N} = \Delta \lambda t  &#10; \Delta \lambda = \dfrac{40}{2000 	imes L} &#10;=0.02

Question 12

In a radioactive decay chain,  _{90}^{232}Th  nucleus decays to  _{82}^{212} Pb  nucleus. Let  N_{\alpha}  and  N_{\beta}  be the number of  \alpha  and  \beta^{-}  particles respectively, emitted in this decay process. Which of the following statements is (are) true?

Accepted Answer

_{90}^{232} Th has converted into   _{82}^{212}Pb &#10;&#10;Change in mass number (A) =  20 &#10;&#10; 	herefore  no of  \alpha  particle   = \dfrac{20}{4} = 5 &#10;&#10;Due to  5 \alpha  particle, z will change by  10  unit.&#10;&#10;Since given change is  8 , therefore no. of the beta particle is  10-8=2

Question 13

Suppose a  ^{226}_{88} Ra  nucleus at rest and in ground state undergoes  \alpha  -decay to a  ^{222}_{86} Rn  nucleus in its excited state. The kinetic energy of the emitted  \alpha  particle is found to be  4.44 MeV .  ^{222}_{86} Rn  nucleus then goes to its ground state by  \gamma -decay. The energy of the emitted  \gamma  photon is ____ keV..[Given  : atomic mass of  ^{226}_{88}Ra = 226.005 u , atomic mass of  ^{222}_{88}Rn = 222.000 u , atomic mass of  \alpha  particle =  4.000 u, \, 1 u = 931 MeV/c^2, c  is speed of light]

Accepted Answer

Mass defect  \Delta m = 226.005 - 222.000 - 4.000 = 0.005 \, amu 	herefore Q   value =  0.005 	imes 931.5 = 4.655 MeV Also  \dfrac{K.E_{\alpha}}{K.E_{Rn}} = \dfrac{m_{Rn}}{m_{\alpha}} \Rightarrow K.E_{Rn} = \dfrac{m_{\alpha}}{m_{Rn}} . K.E_{\alpha} = \dfrac{4}{222} 	imes 4.44 = 0.08 MeV 	herefore  Energy of  \gamma -Photon =  4.655 - (4.44 + 0.08) = 0.135 MeV = 135 KeV

Question 14

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.

Nuclei