Polonium-214 has a relatively short half life of 164 s. How many seconds would it take for 8.0 g of this isotope to decay to 0.25 g?

Question

12923 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-05T07:01:28

$820 \ sec$

$m_i=m_r*2^n$

$m_i " is the initial mass present of the radioisotope."$

$m_r" is the mass remaining after a certain time "$

$n " is the number of periods," = (time)/T" where T is the half-life."$

$m_i=m_r*2^n$

$m_i/m_r=2^n$

$2^n=(8.0\ g)/(0.25\ g)$

$2^n=32$

$2^n=2^5 " " =>$ n = 5

$n= (time)/T$

$time = n xx T$

$time = 5 xx 164 \ sec$

$time = 820 \ sec$

A quick approach:

$underbrace(8\ g -> 4 \ g->2\ g->1\ g ->0.5 \ g-> 0.25 g)$
$" "Time$

$Time = 5 \ periods$ ...