A substance has a half of 3mins. after 6mins. the count rate was observed to be 400. What was its count rate at zero time?
- 200
- 1200
-
1600
- 2400
The correct answer is: C
Explanation
Let the count rate at zero time be K
Count rate after 3mins, = \(\frac{K}{2}\)
Count rate after 6mins = \(\frac{K}{4}\)
but count after 6mins = 400 ( given )
Then = \(\frac{K}{4}\) = 400
K = 4 X 400 = 1600
∴ = count rate at zero time = 1600.