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. `400 mu s `
B. `6.3 mu s `
C. `40 mu s `
D. `300 mu s`

Answer

A

Explanation

`(1)/(16) = (1)/( (t)/(2^(100))) or (1)/(2^(4)) = (1)/(2^(t // 100)) or 4 = (t)/(100) or t = 400 mu s`