A radioactive isotope has h half-life of 2 yr. how long will itt take the activity to reduce to 3% of its original value

Options

A. 48yr
B. 7 yr
C. 10yr
D. 96 yr

Answer

D

Explanation

From `N=N_(0)e^(-lamdat)`
`implies 1800=2000e^(-lamdaxx2)`
`implies (9)/(10)=e^(-2lamda) implies e^(-lamda)=((9)/(10))^(1//2)`
Number of nuclei left after 6s,
`N=N_(0)e^(-lamdat')=2000e^(-lamdaxx6)=2000(e^(-lamda))^(6)`
`=2000[((9)/(10))^(1//2)]^(6)`
`=2000xx((9)/(10))^(3)=2000xx(729)/(1000)=1458`