The half-life of `.^215At` is `100mus`. The time taken for the activity of a sample of `.^215At` to decay to `1/16th` of its initial value is

Options

A. `400mus`
B. `63mus`
C. `40mus`
D. `300mus`

Answer

A

Explanation

`R=R_(0)(1/2)^(n)` …(i)
Here R= acitivity of radioactive substance after n half-lives `=R_(0)/6` (given)
Subtituting in equation (i), we get `n=4`
`:. t=(n)t_(1//2)=(4) (100 mus)=400 mus`