6)
If the observed counting rate of a radioactive sample is 7.4×104
counts per second, then the disintegration rate in Becquerel,
assuming a counting efficiency of 20% will be (1 Becquerel =
2.703×10-11Curie)
(A)
3.7×104
(B)
3.7×105
(C)
4.4×105
(D)
3.7×106
Sol:
$Efficiency = \frac{Observed standard count rate (cps)}{Known standard Disintegration rate(dps)}$
$Disintegration Rate = \frac{Count Rate(cps)}{Efficiency}$
$Efficiency = \frac{Observed standard count rate (cps)}{Known standard Disintegration rate(dps)}$
$Disintegration Rate = \frac{Count Rate(cps)}{Efficiency}$
Efficiency
= 20% =0.2
1
Becquerel = 1 disintegration per second
$Disintegration rate = \frac{7.4\times 10^4}{0.2}=3.7\times 10^4=3.7\times 10^5$
$Disintegration rate = \frac{7.4\times 10^4}{0.2}=3.7\times 10^4=3.7\times 10^5$
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