Question

A 100.0 gram sample of Polonium-210 is contained for 552 days. How many half-lives occur during this period of time, if the half-life is 138 days?

Answer 1

Answer:

4 half-lives will occur during this period of time.

Explanation:

Formula used :

$a_o=a\times e^{-\lambda t}$

$\lambda =\frac{0.693}{t_{1/2}}$

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives and time t

$a_o$ = Initial amount of the reactant.

$\lambda =$ decay constant

$t_{1\2}$ = half life of an isotope

n = number of half lives

We have :

$a_o=100.0 g$

a = ?

t = 552 days

$t_{1/2}=138 days$

$a=100.0 g\times e^{-\frac{0.693}{138}\times 552}$

$a=6.254 g$

$6.254 g=\frac{100.0 g}{2^n}$

$2^n=\frac{100.0 g}{6.254 g}$

n = 4

4 half-lives will occur during this period of time.