After an interval of one day , `1//16th` initial amount of a radioactive material remains in a sample. Then, its half-life( in h) is .

Answer

6

Explanation

`N=(N_(0))/(2^(t//T))`
`(N_(0))/(16) =(N_(0))/(2^(t//T))`
`2^(t//T) =16=2^(4)`
or `T=(t)=(4)`
or `T=(t)/(T)=(24)/(4) h=6h` .