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NEET-XII-Physics

46: The Nucleus

with Solutions - page 6

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  • #19-c
    How much time will it take to convert 25% of this isotope of mercury into gold?
    Ans : Number of active nuclei of mercury at t = 0
    = N0
    = 100
    Active nuclei of mercury left after conversion of 25% isotope of mercury into gold = N = 75
    Now, `` \frac{N}{{N}_{0}}={e}^{-\lambda t}``
    Here, N = Number of inactive nuclei
    N0 = Number of nuclei at t = 0
    `` \lambda `` = Disintegration constant
    On substituting the values, we get
    `` \frac{75}{100}={e}^{-\lambda t}``
    `` \Rightarrow 0.75={\,\mathrm{\,e\,}}^{-\lambda x}``
    `` \Rightarrow \,\mathrm{\,In\,}0.75=-\lambda t``
    `` \Rightarrow t=\frac{\,\mathrm{\,In\,}0.75}{-0.00018}``
    `` =1600\,\mathrm{\,s\,}``
    Page No 443:
  • Qstn #20
    The half-life of 199Au is 2.7 days. (a) Find the activity of a sample containing 1.00 µg of 198Au. (b) What will be the activity after 7 days? Take the atomic weight of 198Au to be 198 g mol-1.
    digAnsr:   b
    Ans : Given:
    Half-life of 199Au, T1/2= 2.7 days
    Disintegration constant, `` \lambda =\frac{0.693}{{T}_{{\displaystyle 1/2}}}`` = `` \frac{0.693}{2.7\times 24\times 60\times 60}=2.97\times {10}^{-6}{\,\mathrm{\,s\,}}^{-1}``
    Number of atoms left undecayed, N = `` \frac{1\times {10}^{-6}\times 6.023\times {10}^{23}}{198}``
    Now, activity, `` {A}_{0}=\lambda N``
    = `` \frac{1\times {10}^{-6}\times 6.023\times {10}^{23}}{198}\times 2.97\times {10}^{-6}``
    = 0.244 Ci
    (b) After 7 days,
    Activity, `` A={A}_{0}{e}^{-\lambda t}``
    `` ``
    Here, A0 = 0.244 Ci
    `` \therefore A=0.244\times {e}^{-2.97\times {10}^{-6}\times 7\times 3600\times 24}``
    `` =0.244\times {e}^{17962.56\times {10}^{-4}}``
    `` =0.040\,\mathrm{\,Ci\,}``
    Page No 443:
  • Qstn #21
    Radioactive 131I has a half-life of 8.0 days. A sample containing 131I has activity 20 µCi at t = 0.
    Ans : Given:
    Half-life of radioactive 131I, T1/2 = 8 days
    Activity of the sample at t = 0, A0 = 20 µCi
    Time, t = 4 days
  • #21-a
    What is its activity at t = 4 days?
    Ans : Disintegration constant, `` \lambda =\frac{0.693}{{T}_{1/2}}=0.0866``
    Activity (A) at t = 4 days is given by
    `` A={A}_{0}{e}^{-\lambda t}``
    `` \Rightarrow A=20\times {10}^{-6}\times {e}^{-0.0866\times 4}``
    `` =1.41\times {10}^{-6}\,\mathrm{\,Ci\,}=14\,\mathrm{\,\mu Ci\,}``
  • #21-b
    What is its decay constant at t = 4.0 days?
    Ans : Decay constant is a constant for a radioactive sample and depends only on its half-life.
    `` \lambda =\frac{0.693}{[8\times 24\times 3600]}``
    `` =1.0026\times {10}^{-6}{\,\mathrm{\,s\,}}^{-1}``
    Page No 443:
  • Qstn #22
    The decay constant of 238U is 4.9 × 10-18 S-1.
    Ans : Given:
    Decay constant, `` \lambda `` = 4.9 × 10-18 s-1
  • #22-a
    What is the average-life of 238U?
    Ans : Average life of uranium `` \left(\tau \right)`` is given by
    `` \tau =\frac{1}{\lambda }``
    `` =\frac{1}{4.9\times {10}^{-18}}``
    `` =\frac{1}{4.9}\times {10}^{18}\,\mathrm{\,s\,}``
    `` =\frac{{10}^{16}}{4.9\times 365\times 24\times 36}\,\mathrm{\,years\,}``
    `` =\frac{{10}^{16}}{4.9\times 365\times 24\times 36}\,\mathrm{\,years\,}``
    `` =6.47\times {10}^{-7}\times {10}^{16}\,\mathrm{\,years\,}``
    `` =6.47\times {10}^{9}\,\mathrm{\,years\,}``
  • #22-b
    What is the half-life of 238U?
    Ans : Half-life of uranium `` \left({T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)`` is given by
    `` {T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{\lambda }=\frac{0.693}{4.9\times {10}^{-18}}``
    `` =\frac{0.693}{4.9}\times {10}^{18}\,\mathrm{\,s\,}``
    `` =0.1414\times {10}^{18}\,\mathrm{\,s\,}``
    `` =\frac{0.1414\times {10}^{18}}{365\times 24\times 3600}``
    `` =\frac{1414\times {10}^{12}}{365\times 24\times 36}``
    `` =4.48\times {10}^{-3}\times {10}^{12}``
    `` =4.5\times {10}^{9}\,\mathrm{\,years\,}``
    `` ``
  • #22-c
    By what factor does the activity of a 238U sample decrease in 9 × 109 years?
    Ans : Time, t = 9 × 109 years
    Activity (A) of the sample, at any time t, is given by
    `` A=\frac{{A}_{0}}{{\displaystyle {2}^{\frac{t}{{T}_{1/2}}}}}``
    `` ``
    Here, A0 = Activity of the sample at t = 0
    `` \therefore \frac{{A}_{0}}{A}={2}^{\frac{9\times {10}^{9}}{4.5\times {10}^{9}}}={2}^{2}=4``
    Page No 443:
  • Qstn #23
    A certain sample of a radioactive material decays at the rate of 500 per second at a certain time. The count rate falls to 200 per second after 50 minutes.
    Ans : Given:
    Initial rate of decay, A0 = 500,
    Rate of decay after 50 minutes, A = 200
    Time, t = 50 min
    = 50 `` \times `` 60
    = 3000 s
  • #23-a
    What is the decay constant of the sample?
    Ans : Activity, A = A0e-λt
    Here, `` \lambda `` = Disintegration constant
    `` \therefore `` 200 = 500 × e-50×60×λ
    `` \Rightarrow \frac{2}{5}={e}^{-3000\lambda }``
    `` \Rightarrow \,\mathrm{\,In\,}\frac{2}{5}=-3000\lambda ``
    `` \Rightarrow \lambda =3.05\times {10}^{-4}{\,\mathrm{\,s\,}}^{-1}``
    `` ``
  • #23-b
    What is its half-life?
    Ans : `` \,\mathrm{\,Half\,}-\,\mathrm{\,life\,},{T}_{1/2}=\frac{0.693}{\lambda }``
    `` =2272.13\,\mathrm{\,s\,}``
    `` =38\,\mathrm{\,min\,}``
    Page No 443:
  • Qstn #24
    The count rate from a radioactive sample falls from 4.0 × 106 per second to 1.0 × 106 per second in 20 hours. What will be the count rate 100 hours after the beginning?
    Ans : Given:
    Initial count rate of radioactive sample, A0 = 4 × 106 disintegration/sec
    Count rate of radioactive sample after 20 hours, A = 1 × 106 disintegration/sec
    Time, t = 20 hours
    Activity of radioactive sample `` \left(A\right)`` is given by
    `` A=\frac{{A}_{0}}{{2}^{{\displaystyle \frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}}``
    `` \,\mathrm{\,Here\,},{T}_{1/2}=\,\mathrm{\,Half\,}-\,\mathrm{\,life\,}\,\mathrm{\,period\,}``
    `` \,\mathrm{\,On\,}\,\mathrm{\,substituting\,}\,\mathrm{\,the\,}\,\mathrm{\,values\,}\,\mathrm{\,of\,}{A}_{0}\,\mathrm{\,and\,}A,\,\mathrm{\,we\,}\,\mathrm{\,have\,}``
    `` {2}^{\frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}={2}^{2}``
    `` \Rightarrow \frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=2``
    `` \Rightarrow \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=t/2=20\,\mathrm{\,h\,}/2=10\,\mathrm{\,h\,}``
    `` ``
    100 hours after the beginning,
    `` \,\mathrm{\,Count\,}\,\mathrm{\,rate\,},A"=\frac{{A}_{0}}{{2}^{{\displaystyle \frac{t}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}}``
    `` \Rightarrow A"=\frac{4\times {10}^{6}}{{2}^{100/10}}``
    `` =0.390625\times {10}^{4}``
    `` =3.9\times {10}^{3}\,\mathrm{\,disintegrations\,}/\,\mathrm{\,sec\,}``
    Page No 443:
  • Qstn #25
    The half-life of 226Ra is 1602 y. Calculate the activity of 0.1 g of RaCl2 in which all the radium is in the form of 226Ra. Taken atomic weight of Ra to be 226 g mol-1 and that of Cl to be 35.5 g mol-1.
    Ans : Given:
    Half-life of radium, T1/2 = 1602 years
    Atomic weight of radium = 226 g/mole
    Atomic weight of chlorine = 35.5 g/mole
    Now,
    1 mole of RaCl2 = 226 + 71 = 297 g
    297 g = 1 mole of RaCl2
    0.1 g = `` \frac{1}{297}\times 0.1\,\mathrm{\,mole\,}\,\mathrm{\,of\,}{\,\mathrm{\,RaCl\,}}_{2}``
    Total number of atoms in 0.1 g of RaCl2, N `` =\frac{0.1\times 6.023\times {10}^{23}}{297}=0.02027\times {10}^{22}``
    `` ``
    ∴ No of atoms, N = 0.02027 `` \times `` 1022
    `` \,\mathrm{\,Disintegration\,}\,\mathrm{\,constant\,},\lambda =\frac{0.693}{{T}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}``
    `` =\frac{0.693}{1602\times 365\times 24\times 3600}``
    `` =1.371\times {10}^{-11}``
    Activity of radioactive sample, A = `` \lambda N``
    = `` 1.371\times {10}^{-11}\times 2.027\times {10}^{20}``
    = `` 2.8\times {10}^{9}\,\mathrm{\,disintegrations\,}/\,\mathrm{\,second\,}``
    Page No 443:
  • Qstn #26
    The half-life of a radioisotope is 10 h. Find the total number of disintegration in the tenth hour measured from a time when the activity was 1 Ci.
    Ans : Given:
    Half-life of radioisotope, `` {T}_{1/2}`` = 10 hrs
    Initial activity, A0 = 1 Ci
    Disintegration constant, `` \lambda =\frac{0.693}{10\times 3600}{\,\mathrm{\,s\,}}^{-1}``
    Activity of radioactive sample, `` A={A}_{0}{e}^{-\lambda t}``
    `` ``</math><br><math xmlns="http://www.w3.org/1998/Math/MathML"></math><br><math xmlns="http://www.w3.org/1998/Math/MathML">
    Page No 443:
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