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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
The Nucleus

Concepts Of Physics MCQ Edition [Volume 2] · Physics · The Nucleus

Determine the activity of a sample containing initially 1.00 of ^...

Determine the activity of a sample containing initially 1.00μg1.00 \mu\text{g} of ^{198}\text{Au}afterafter7 \text{ days}haveelapsed.Thehalf−lifeof198Auhave elapsed. The half-life of ^{198}\text{Au} is 2.7 days2.7 \text{ days} and its atomic weight is 198 g mol−1198 \text{ g mol}^{-1}.

Options

  1. A0.010 Ci0.010 \text{ Ci}
  2. B0.080 Ci0.080 \text{ Ci}
  3. C0.020 Ci0.020 \text{ Ci}
  4. D0.040 Ci0.040 \text{ Ci}

Correct answer

D. 0.040 Ci0.040 \text{ Ci}

Step-by-step solution

The initial number of atoms N0N_0 in the sample is given by:

N0=mM×NAN_0 = \dfrac{m}{M} \times N_A

Substituting m=1.00μg=10−6 gm = 1.00 \mu\text{g} = 10^{-6} \text{ g}, M=198 g mol−1M = 198 \text{ g mol}^{-1}, and NA=6.023×1023 mol−1N_A = 6.023 \times 10^{23} \text{ mol}^{-1}:

N0=10−6198×6.023×1023≈3.04×1015 atomsN_0 = \dfrac{10^{-6}}{198} \times 6.023 \times 10^{23} \approx 3.04 \times 10^{15} \text{ atoms}

The decay constant λ\lambda is:

λ=ln⁡2T1/2=0.6932.7×24×3600≈2.97×10−6 s−1\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{2.7 \times 24 \times 3600} \approx 2.97 \times 10^{-6} \text{ s}^{-1}

The initial activity A0A_0 is:

A0=λN0=(2.97×10−6)×(3.04×1015)≈9.03×109 decays/sA_0 = \lambda N_0 = (2.97 \times 10^{-6}) \times (3.04 \times 10^{15}) \approx 9.03 \times 10^9 \text{ decays/s}

Converting this to Curies (1 Ci=3.7×1010 decays/s1 \text{ Ci} = 3.7 \times 10^{10} \text{ decays/s}):

A0=9.03×1093.7×1010≈0.244 CiA_0 = \dfrac{9.03 \times 10^9}{3.7 \times 10^{10}} \approx 0.244 \text{ Ci}

The activity AA after t=7 dayst = 7 \text{ days} is:

A=A0(12)tT1/2A = A_0 \left(\dfrac{1}{2}\right)^{\dfrac{t}{T_{1/2}}}

A=0.244×(12)72.7A = 0.244 \times \left(\dfrac{1}{2}\right)^{\dfrac{7}{2.7}}

A=0.244×(12)2.59≈0.244×0.166≈0.040 CiA = 0.244 \times \left(\dfrac{1}{2}\right)^{2.59} \approx 0.244 \times 0.166 \approx 0.040 \text{ Ci}
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