In 5.49 seconds, 1.20 g of argon-35 decays to leave only .15 g. What is the half life of argon-35?

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Answer 1 · 2016-05-22T07:49:58

1.83 sec

$m_i=m_r*2^n$

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

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

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

$m_i=m_r*2^n$

$m_i/m_r=2^n$

$2^n=(1.20\ g)/(0.15\ g)$

$2^n=8$

$2^n=2^3$

$" "color(red)ulbar(color(black)(n=3))$

$n= (time)/T"$

$T "(half-life)"= (time)/n$

$T = (5.49 sec)/3$

$color(blue)(ulbar(color(red)(T = 1.83" sec"))$