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NEET-XII-Physics

46: The Nucleus

with Solutions - page 5
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  • #19
    The decay constant of
    Hg80197(electron capture to
    Au79197) is 1.8 × 10-4 S-1. (a) What is the half-life? (b) What is the average-life? (c) How much time will it take to convert 25% of this isotope of mercury into gold? (a) What is the half-life? (b) What is the average-life? (c) How much time will it take to convert 25% of this isotope of mercury into gold?
    Ans : Given:
    Decay constant of `` {}_{80}{}^{197}\,\mathrm{\,Hg\,}``, `` \lambda `` = 1.8 × 10-4 s`` -1`` (a) Half-life, `` {T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{\lambda }``
    `` \Rightarrow {T}_{1/2}=\frac{0.693}{1.8\times {10}^{-4}}``
    `` =3850\,\mathrm{\,s\,}=64\,\mathrm{\,minutes\,}``
    `` `` (b) `` \,\mathrm{\,Average\,}\,\mathrm{\,life\,},{T}_{av}=\frac{{T}_{1/2}}{0.693}``
    `` =\frac{64}{0.693}``
    `` =92\,\mathrm{\,minutes\,}``
    `` `` (c) 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: (a) Half-life, `` {T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{\lambda }``
    `` \Rightarrow {T}_{1/2}=\frac{0.693}{1.8\times {10}^{-4}}``
    `` =3850\,\mathrm{\,s\,}=64\,\mathrm{\,minutes\,}``
    `` `` (b) `` \,\mathrm{\,Average\,}\,\mathrm{\,life\,},{T}_{av}=\frac{{T}_{1/2}}{0.693}``
    `` =\frac{64}{0.693}``
    `` =92\,\mathrm{\,minutes\,}``
    `` `` (c) 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:
  • #19-a
    What is the half-life?
    Ans : Half-life, `` {T}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{0.693}{\lambda }``
    `` \Rightarrow {T}_{1/2}=\frac{0.693}{1.8\times {10}^{-4}}``
    `` =3850\,\mathrm{\,s\,}=64\,\mathrm{\,minutes\,}``
    `` ``
  • #19-b
    What is the average-life?
    Ans : `` \,\mathrm{\,Average\,}\,\mathrm{\,life\,},{T}_{av}=\frac{{T}_{1/2}}{0.693}``
    `` =\frac{64}{0.693}``
    `` =92\,\mathrm{\,minutes\,}``
    `` ``
  • #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: