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`.

Options

A. 3737 y
B. 8327 y
C. 7823 y
D. 2378 y

Answer

D

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 of time t older than the sample A.
We have, `A = A_(0) e^(-lambda t)`
or `1400 s^(-1) = 2100 s^(-1) e^(-lambda t) or e^(-lambda t) = (2)/(3)`
`therefore" "t = (1n(3//2))/(lambda) = (1n (3//2))/(0.693) t_(1//2)`
`= (0.4055)/(0.693) xx 5730 y = 3352 y`
So, the age difference between two samples is, t' = 5730 - 3352 = 2378 y.