# After 42 days a 2.0 g sample of phosphorus-32 contains only 0.25 g of the isotope. What is the half-life of phosphorus-32?

May 13, 2018

14 days

#### Explanation:

Mass of atoms initially,${N}_{0}$=2.0 g
Mass after 42 days,$N$=0.25 g $i . e$ mass left undecayed
we have,
$\frac{N}{N} _ 0 = \frac{1}{2} ^ n$, where $n$ is number of half lives occured(the number of times it disintegrated into half of original)
$\implies \frac{0.25}{2} = \frac{1}{2} ^ n$
$\implies \frac{1}{8} = \frac{1}{2} ^ n$
$\implies n = 3$
$\implies$number of half lives occured in 42 days is 3
if half life is ${t}_{\frac{1}{2}}$
then $3 {t}_{\frac{1}{2}} = 42$
or ${t}_{\frac{1}{2}} = 14$
$\therefore$half life is of 14 days

May 13, 2018

The half life is $= 14 \text{ days }$

#### Explanation:

The proportion of $\text{ Phosphorus- 32 }$ remaining after $42$ days is

$= {m}_{t} / {m}_{0} = \frac{0.25}{2} = \frac{1}{8}$

This represents

$= 3$ half lives

$42$ days represent $3 \cdot {t}_{\frac{1}{2}}$

Therefore,

$3 \cdot {t}_{\frac{1}{2}} = 42$

${t}_{\frac{1}{2}} = \frac{42}{3} = 14 \text{ days }$

The half life is ${t}_{\frac{1}{2}} = 14 \text{days}$