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. `2xx10^(10)`
B. `1.5xx10^(10)`
C. zero
D. infinity

Answer

B

Explanation

`As,N=N_(0)((1)/(2))^(t//T)` Number of atoms at t = 2 h,
`N_(1)=8xx10^(10)((1)/(2))^(2/1)=2xx10^(10)`
Number of atoms at t = 4 h,
`N_(2)=8xx10^(10)((1)/(2))^(4/1)=1/2xx10^(10)`
`therefore` Number of atoms decayed in given duration
`=(2-(1)/(2))xx10^(10)=1.5xx10^(10)`