The radioactive decay rate of a radioactive element is found to be `10^(3) ` disintegration per second at a cartain time . If the half life of the element is one second , the dacay rate after one second ….. And after three second is ……

Answer

b

Explanation

` (dN)/(dt)=100` at `t =0 s`
`t_(1//2) =1 s rArr lambda =0. 692 s^(-1)
At `t=1 s`,
`1 =(2.303)/(0.693) log 10 (N_(0))/(N)`
now,
`t=2.302)/(lambda) log_(10). (N_(0))/(N)`
:. 0.3010 =log_(10) .(N_(0))/(N) rArr gol_(10)_(N_(0))/(N) =log_(10^(2))`
`:.(N_(0))/(N)=2rArr N=(N_(0))/(N)`
`:. N=(100)/(2)=500`
Here,
`(dN)/(dt) prop N_(0)`
and `(dN)/(dt) prop
`:. (dN)/(dt)=500 dps ` at `t=1 s`
Similarly, at `t=3 s`, (dN)/(dt)=125 dps`.
Alternative solution :
`A=A_(0)((1)/(2))^(n)`
Where `A_(0)` =initial activity =100 dps(given), A=activity after n half-lives.
At `t=1,n=1 rArr A=1000(1/2)^(1)=500 dps`
At `t=3, n=3 rArr A=1000(1/2)^(3)=125 dps`