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

Answer:

a) $\frac{N}{N_{0}}=0.1353$

b) $\frac{N}{N_{0}}=0.0183$

c) $\frac{N}{N_{0}}=4.54\times 10^{-5}$

Explanation:

using equation

$N=N_{0}e^{(-\frac{t}{T} )}$

where T is  half lives

a) two half-lives

$N=N_{0}e^{(-\frac{t}{T} )}$

$N=N_{0}e^{(-\frac{2T}{T} )}$

$\frac{N}{N_{0}}=e^{(-\frac{2T}{T} )}$

$\frac{N}{N_{0}}=0.1353$

b) four half-lives

$N=N_{0}e^{(-\frac{t}{T} )}$

$\frac{N}{N_{0}}=e^{(-\frac{4T}{T} )}$

$\frac{N}{N_{0}}=0.0183$

c) 10 half-lives

$N=N_{0}e^{(-\frac{t}{T} )}$

$\frac{N}{N_{0}}=e^{(-\frac{10T}{T} )}$

$\frac{N}{N_{0}}=4.54\times 10^{-5}$