Equal masses of two samples of charcoal `A and B` are burnt separately and the resulting carbon dioxide are collected in two vessels. The radioactivity of `^14 C` is measured for both the gas samples. The gas from the charcoal A gives `2100 counts` per week and the gas from the charcoal A gives 2100 counts per week and the gas from the charcoal `B` gives `1400` counts per week. Find the age difference between the two samples. Half-life of `^14 C = 5730 y`.

Explanation

The activity of sample `A` is 2100 counts per week. After a certain time `t`, its activity will be reduced to 1400 counts per week. This is because a fraction of the active `^14 C` nuclei will decay in time `t`. The sample `B` must be a time `t` older than the sample `A`.
We have,
`A = A_0 e^(-lambdat)`
or, `1400 s^(-1) = 2100 s^(-1) e^(-lambdat)`
or, `e^(-lambdat) = 2/3`
`t = (1n(3/2))/(lambda)`
`= (1n(3/2))/(0.693) t_(1/2)`
`= (0.4050)/(0.693) xx 5730 y = 3352 y`.