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CBSE Class 12 Physics 2020 Delhi Set 1 Paper

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Question : 30 of 37
Marks: +1, -0
(a) Define the term decay constant of a radioactive substance.
(b) The half-life of 92238U{}_{92}^{238}\mathrm{U} undergoing a decay is 4.5×1094.5 \times 10^9 years. Calculate the activity of 10g10\,\mathrm{g} sample of 92238U{}_{92}^{238}\mathrm{U}.
Solution:  
(a) Decay constant: Let the number of nuclei of a radioactive substance present at any time be NN. Decay of number of nuclei (ΔN)(\Delta N) in a small interval of time (Δt)(\Delta t) is proportional to NN and Δt\Delta t.
  ΔNNΔt\;\Delta N \propto N \Delta t
   Or,     ΔN=λNΔt\;\text{ Or, }\;\;\Delta N = -\lambda N \Delta t
The proportionality constant λ\lambda is known as the decay constant.
(b) Half-life =4.5×109=4.5 \times 10^9 years
Mass number =238=238
  λ=0.693T=0.6934.5×109   per year   \;\lambda = \frac{0.693}{T} = \frac{0.693}{4.5 \times 10^9} \;\text{ per year }\;
  N=10×6.02×1023238\; N = \frac{10 \times 6.02 \times 10^{23}}{238}
Hence, the activity, A=λNA = \lambda N
     Or,     A=0.6934.5×109×10×6.02×1023238\;\;\text{ Or, }\;\; A = \frac{0.693}{4.5 \times 10^9} \times \frac{10 \times 6.02 \times 10^{23}}{238}
  A=0.0039×1015=3.9×1012   per year   \; \therefore A = 0.0039 \times 10^{15} = 3.9 \times 10^{12} \;\text{ per year }\;
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