The half-life of a certain radioactive specimen is to be measured. We have a Geiger Mueller counter which measures the activity rate. We start the counter. At t=0, the activity rate is 1000+- 5 dps. At t=100 sec, the activity drops to `500 +- 5` dps. If the least count (L.C) of stop watch is 1 sec, find the percentage error in the half life.

Options

A. 1.43
B. 2.16
C. 3.32
D. 4.04

Answer

B

Explanation

`A_(t)= A_(0)(0.5)^(t//t_(1//2))`
`(1/2)^(1) = (1/2)^(t/t_(1//2)) therefore t=t_(1//2)`
`A_(t) = A_(0)e^(-lambda t)`
In `A_(t)= In A_(0) - lambda t`
`(ln 2)/(t_(1//2)) = lambda t = ln A_(0) - ln A_(t)`
`(ln 2)/(-(t_(1//2))^(2) dt_(1//2) t) = (dA_(0))/A_(0) + (dA_(t))/A`
`(dt_(1//2))/t_(1//2) = ((dA_(0))/A_(0) + (dA_(t))/A_(t))/(ln2) = (5/1000 + 5/500)/(ln 2) = 1.5/(ln 2)`