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?

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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?

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Answer 1 · 2018-05-13T08:19:09

14 days

Explanation:

Mass of atoms initially, $N_{0}$ $N_0$ =2.0 g
Mass after 42 days, $N$ $N$ =0.25 g $i . e$ $i.e$ mass left undecayed
we have,
$\frac{N}{N_{0}} = \frac{1}{2^{n}}$ $N/N_0=1/2^n$ , where $n$ $n$ is number of half lives occured(the number of times it disintegrated into half of original)
$\Rightarrow \frac{0.25}{2} = \frac{1}{2^{n}}$ $=>0.25/2=1/2^n$
$\Rightarrow \frac{1}{8} = \frac{1}{2^{n}}$ $=>1/8=1/2^n$
$\Rightarrow n = 3$ $=>n=3$
$\Rightarrow$ $=>$ number of half lives occured in 42 days is 3
if half life is $t_{\frac{1}{2}}$ $t_(1/2)$
then $3 t_{\frac{1}{2}} = 42$ $3t_(1/2)=42$
or $t_{\frac{1}{2}} = 14$ $t_(1/2)=14$
$:.$ half life is of 14 days

Answer 2 · 2018-05-13T08:23:50

The half life is $=14 " days "$

Explanation:

![http://chemwiki.ucdavis.edu](https://useruploads.socratic.org/i0nzxwQiR1uX7T9ILCxu_Radioactive_Decay.jpg)

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 $t_(1/2)=14 "days"$

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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?

2 Answers

Explanation:

Explanation:

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