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 undergoing a decay is 4.5×109 years. Calculate the activity of 10g sample of 92238U.
Solution:
(a) Decay constant: Let the number of nuclei of a radioactive substance present at any time be N. Decay of number of nuclei (N) in a small interval of time (t) is proportional to N and t.
NNt
Or, N=λNt
The proportionality constant λ is known as the decay constant.
(b) Half-life =4.5×109 years
Mass number =238
λ=0.693T=0.6934.5×109 per year
N=10×6.02×1023238
Hence, the activity, A=λN
Or, A=0.6934.5×109×10×6.02×1023238
A=0.0039×1015=3.9×1012 per year
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