The half-life of a sample of a radioactive substance is `1` hour. If `8 xx 10^10` atoms are present at `t = 0`, then the number of atoms decayed in the duration `t = 2` hour to `t = 4` hour will be

Options

A. `2 xx 10^10`
B. `1.5 xx 10^10`
C. Zero
D. Infinity

Answer

B

Explanation

(b) `N = N_0 ((1)/(2))^((t)/(T_1//2))`
No of atoms at `t = 2hr`,
`N_1 = 8 xx 10^10 ((1)/(2))^((2)/(1)) = 2 xx 10^10`
No. of atoms at `t = 4 hr`,
`N_1 = 8 xx 10^10 ((1)/(2))^((4)/(1)) = (1)/(2) xx 10^10`
`:.` No. pf atoms decayed in given duration
=`(2 - (1)/(2)) xx 10^10 = 1.5 xx 10^10`.