An active nucleus decays to `((1)/(3))` rd in 20 h.
The fraction of original activity remaining after 80 h is

Options

A. `(1)/(16)`
B. `(1)/(81)`
C. `(1)/(36)`
D. `(1)/(54)`

Answer

B

Explanation

Decay equation, `N=N_(0)" exp"(-lambda t)`
Given, `" "N_(1)=N_(0)//3,N_(2)=?`
`t_(1)=20h" "t_(2)=20" h"`
So, `" "N_(1)=N_(0)" exp"(-lambda t_(1))`
`implies" "(1)/(3)=e(-20lambda)impliese^(20lambda)=3`
Now, `N_(2)=N_(0)" exp"(-lambda t_(2))implies(N_(0))/(N_(2))=e^(80 lambda)`
`implies" "(N_(0))/(N_(2))=(e^(20lambda))^(4)=(3)^(4)`
`implies" "N_(2)=(1)/((3)^(4))N_(0)=(N_(0))/(81)`