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CBSE Class 12-science Answered

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Asked by sahupramod650 | 19 Nov, 2019, 10:36: AM
answered-by-expert Expert Answer
Relation involving half life T1/2  and decay constant λ is  given by,   λ =0.693 / T1/2  .............(1)
 
hence decay constant λ for 238U92 = 0.693 /( 4.5 × 109 × 365 × 24 × 3600 ) = 4.883 × 10-18 s-1
 
activity , dN/dt = λN ...............(2)
 
where N is number of atoms in 1 gm Uranium,   N = 6.02 × 1023 /  238 = 2.529 × 1021
 
activity = 4.883 × 10-18 × 2.529 × 1021 = 12.35 × 103 Bq = 12350 Bq
 
( 1 Bq = i disintegration per sec )
 
activity in Curie = 12350/(3.7 × 1010 ) = 3.34 × 10-7 Ci  = 0.334 μCi
Answered by Thiyagarajan K | 19 Nov, 2019, 12:24: PM

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