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November 28, 2014

November 28, 2014

Posted by **Ryan** on Thursday, December 13, 2007 at 9:07pm.

the problem is:

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

- chemistry -
**Damon**, Thursday, December 13, 2007 at 9:29pmmass at t = mass at beginning times e^-t/T

where T is some decay time

m = mo e^-t/T

now when will m = 1/2 mo ?

1/2 = e^- thalf/T where I am calling thalf the half life

-ln (1/2) = thalf/T

but -ln(1/2) = .6931

so

T = thalf/.6931

NOW for this problem

.25 = 2 e^-42/T

ln (.125) = -42/T

T = 20.2 days

so

thalf = .6931(20.2 = 14 days

rough check

after 14 days I have (1/2)2 = 1 gram

after 28 days I have (1/2)1 = .5 gram

after 42 days I have (1/2)(.5)=.25 grams

well, amazing, it worked :)

- chemistry -
**DrBob222**, Thursday, December 13, 2007 at 10:56pmThe same math but in slightly different format.

k=0.693/t_{1/2}

ln(No/N) = kt where No = beginning grams, N = ending grams, k is the constant from equation 1 and t is the time.

substitute from equation 1 into k in equation 2 to obtain

ln(No/N) = [0.693/t_{1/2}*t]

Now substitute your numbers

ln(2/0.25) = [0.693/t_{1/2}*42]

ln 8 = [0.693*42/t_{1/2}]

2.079 = 29.1/t_{1/2}

t_{1/2}= 29.1/2.079 = 13.997 which rounds to 14 days.

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