NEET Class neet Answered
please answer this
Asked by Prashant DIGHE | 14 Jan, 2020, 10:11: PM
Expert Answer
Half life of radioactive element = 20 min ; decay constant λ = ( ln2 ) /20 ≈ 0.0347 min-1
Let after t1 seconds, 33% is disintegrated , hence we have 0.67 = e-λ × t1 .........................(1)
Let after t2 seconds, 67% is disintegrated , hence we have 0.33 = e-λ × t2 .........................(2)
By dividing eqn.(2) with eqn.(1), we get, e-λ × ( t2 - t1 ) = 0.33 / 0.67 ≈ 0.5
Hence , λ ( t2 - t1 ) = ln2 or (t2-t1) = (ln2 ) / λ = ( ln2 ) /0.0347 ≈ 20 min
Answered by Thiyagarajan K | 15 Jan, 2020, 08:02: AM
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