Half-life of a radioactive substance is 18 minutes. The time interval between its 20% decay and 80% decay in minutes is

Options

A. 6
B. 9
C. 18
D. 36

Answer

D

Explanation

After n half - life, the numbers of atom left undecayed is given by,
`N = N_(0) ((1)/(2))^(n)`
`therefore " " n = (t )/(T_((1)/(2))) or N = N_(0) ((1)/(2))^((t)/(T_(i)//2)) `
For 20% decay of radioactive substance,
N = `N_(0) - (20)/(100) N_(0) = 0.8 N_(0)`
Or `" " 0.8 N_(0) = N_(0) ((1)/(2))^(t_(1)//18) " " ` ... (i )
For 80% decay of radioactive substance,
N = `N_(0) - (80)/(100) N_(0) = 0.2 N_(0) or 0.2 N_(0) = N_(0) ((1)/(2))^(t_(2)//18) `
From Eqs. (i) and (ii), we get `rArr 4 = ((1)/(2))^((t_(1) - t_(2))/(18))`
Taking log on the both sides, we get
log 4 = log ` ((1)/(2))^((t_(1) - t_(2))/(18)) rArr log 2^(2) = log 2^(((t_(2) - t_(1))/(18))`
`rArr " " 2 = (t_(2) - t_(1))/(18) " " rArr t_(2) - t_(1) = 36 ` min