Half-life of a radioacitve element is `10` days. The time during which quantity remains `1//10` of initial mass will be

Options

A. 100 days
B. 33 days
C. 25 days
D. 48 days

Explanation

`N=N_(0)^(-lamdat)=In(N_(s))/N=lamdat`
`t=1/(lamda)In(N_(s))/Nimpliest=(2.303xxT_(1//2))log_(30)(N_(s))/N`
`(N_(s))/N=10,T_(1//2)=10` days
`:.t=(2.303xx10)/0.693=33.23` days