Question

What fraction of the original number of nuclei present in a sample will remain after two half-lives; four half-lives; 10 half-lives?

Answer 1

a)
`(N)/(N_(0))=0.1353`

b)
`(N)/(N_(0))=0.0183`

c)
`(N)/(N_(0))=4.54* 10^(-5)`

Step-by-step explanation:

using equation

`N=N_(0)e^{(-(t)/(T) )}`

where T is half lives

a) two half-lives

`N=N_(0)e^{(-(t)/(T) )}`

`N=N_(0)e^{(-(2T)/(T) )}`

`(N)/(N_(0))=e^{(-(2T)/(T) )}`

`(N)/(N_(0))=0.1353`

b) four half-lives

`N=N_(0)e^{(-(t)/(T) )}`

`(N)/(N_(0))=e^{(-(4T)/(T) )}`

`(N)/(N_(0))=0.0183`

c) 10 half-lives

`N=N_(0)e^{(-(t)/(T) )}`

`(N)/(N_(0))=e^{(-(10T)/(T) )}`

`(N)/(N_(0))=4.54* 10^(-5)`