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NEET-XII-Physics

46: The Nucleus

with Solutions - page 6
Qstn# iv-22-a Prvs-QstnNext-Qstn
  • #22-a
    What is the average-life of 238U? (b) What is the half-life of 238U? (c) By what factor does the activity of a 238U sample decrease in 9 × 109 years? (b) What is the half-life of 238U? (c) By what factor does the activity of a 238U sample decrease in 9 × 109 years?
    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\,}`` (b) 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\,}``
    `` `` (c) 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: (b) 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\,}``
    `` `` (c) 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:
  • #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: