Question

# A radioactive sample at any instant has its disintegration rate 5000 disintegration per minute. After 5 minutes, the rate is 1250 disintegrations per minute. Then, the decay constant, per minute, is

A
0.4ln2
B
0.2ln2
C
0.1ln2
D
0.8ln2
Solution
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#### Let decay constant per minute=λTherefore, ATQ, initially disintegration rate,R=N∘λ=5000→(1)Also, in 2nd case, disintegration rate after 5 min is,$1250=N\lambda\rightarrow (2)$Dividing equation (1) by (2), we getN∘λNλ=60001250N∘N=4And as we know, the number of decay, N is given by,N=N∘e−λt(t=5min)14=e−5λ−ln(4)=−5λλ=0.4ln2

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