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NEET-XII-Physics

46: The Nucleus

with Solutions - page 7
Qstn# iv-35 Prvs-QstnNext-Qstn
  • #35
    The count rate of nuclear radiation coming from a radiation coming from a radioactive sample containing 128I varies with time as follows.


    Time t (minute): 0 25 50 75 100
    Ctount rate R (109 s-1): 30 16 8.0 3.8 2.0
    (a) Plot In (R0/R) against t. (b) From the slope of the best straight line through the points, find the decay constant λ. (c) Calculate the half-life t1/2.
    Ans : (a) For t = 0,
    `` \,\mathrm{\,ln\,}\left(\frac{{R}_{\mathit{0}}}{R}\right)=\,\mathrm{\,In\,}\left(\frac{30\times {10}^{9}}{30\times {10}^{9}}\right)=0``
    `` ``
    For t = 25 s,
    `` \,\mathrm{\,ln\,}\left(\frac{{R}_{\mathit{0}}}{{R}_{\mathit{2}}}\right)=\,\mathrm{\,In\,}\left(\frac{30\times {10}^{9}}{16\times {10}^{9}}\right)=0.63``
    For t = 50 s,
    `` \,\mathrm{\,In\,}\left(\frac{{\mathit{R}}_{\mathit{0}}}{{\mathit{R}}_{\mathit{3}}}\right)=\,\mathrm{\,In\,}\left(\frac{30\times {10}^{9}}{8\times {10}^{9}}\right)=1.35``
    For t = 75 s,
    `` \,\mathrm{\,ln\,}\left(\frac{{R}_{0}}{{R}_{4}}\right)=\,\mathrm{\,In\,}\left(\frac{30\times {10}^{9}}{3.8\times {10}^{9}}\right)=2.06``
    For t = 100 s,
    `` \,\mathrm{\,In\,}\left(\frac{{R}_{0}}{{R}_{5}}\right)=\,\mathrm{\,In\,}\left(\frac{30\times {10}^{9}}{2\times {10}^{9}}\right)=2.7``
    The required graph is shown below.
    (b) Slope of the graph = 0.028
    ∴ Decay constant, `` \lambda `` = 0.028 min`` -1``
    The half-life period`` \left({T}_{\frac{1}{2}}\right)`` is given by
    `` {T}_{\frac{1}{2}}=\frac{0.693}{\lambda }``
    `` =\frac{0.693}{0.028}=25\,\mathrm{\,min\,}``
    Page No 443:
  • #35-a
    Plot In (R0/R) against t.
    Ans : For t = 0,
    `` \,\mathrm{\,ln\,}\left(\frac{{R}_{\mathit{0}}}{R}\right)=\,\mathrm{\,In\,}\left(\frac{30\times {10}^{9}}{30\times {10}^{9}}\right)=0``
    `` ``
    For t = 25 s,
    `` \,\mathrm{\,ln\,}\left(\frac{{R}_{\mathit{0}}}{{R}_{\mathit{2}}}\right)=\,\mathrm{\,In\,}\left(\frac{30\times {10}^{9}}{16\times {10}^{9}}\right)=0.63``
    For t = 50 s,
    `` \,\mathrm{\,In\,}\left(\frac{{\mathit{R}}_{\mathit{0}}}{{\mathit{R}}_{\mathit{3}}}\right)=\,\mathrm{\,In\,}\left(\frac{30\times {10}^{9}}{8\times {10}^{9}}\right)=1.35``
    For t = 75 s,
    `` \,\mathrm{\,ln\,}\left(\frac{{R}_{0}}{{R}_{4}}\right)=\,\mathrm{\,In\,}\left(\frac{30\times {10}^{9}}{3.8\times {10}^{9}}\right)=2.06``
    For t = 100 s,
    `` \,\mathrm{\,In\,}\left(\frac{{R}_{0}}{{R}_{5}}\right)=\,\mathrm{\,In\,}\left(\frac{30\times {10}^{9}}{2\times {10}^{9}}\right)=2.7``
    The required graph is shown below.
  • #35-b
    From the slope of the best straight line through the points, find the decay constant λ. (c) Calculate the half-life t1/2.
    Ans : Slope of the graph = 0.028
    ∴ Decay constant, `` \lambda `` = 0.028 min`` -1``
    The half-life period`` \left({T}_{\frac{1}{2}}\right)`` is given by
    `` {T}_{\frac{1}{2}}=\frac{0.693}{\lambda }``
    `` =\frac{0.693}{0.028}=25\,\mathrm{\,min\,}``
    Page No 443: