Question

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?

Answer 1

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,
`N/N_0=1/2^n`, where `n` is number of half lives occured(the number of times it disintegrated into half of original)
`=>0.25/2=1/2^n`
`=>1/8=1/2^n`
`=>n=3`
`=>`number of half lives occured in 42 days is 3
if half life is `t_(1/2)`
then `3t_(1/2)=42`
or `t_(1/2)=14`
`:.`half life is of 14 days

Answer 2

The half life is `=14 " days "`

Explanation:

The proportion of `" Phosphorus- 32 "` remaining after `42` days is

`=m_t/m_0=0.25/2=1/8`

This represents

`=3` half lives

`42` days represent `3*t_(1/2)`

Therefore,

`3*t_(1/2)=42`

`t_(1/2)=42/3=14 " days "`

The half life is `t_(1/2)=14 "days"`