If 12.5 grams of the original sample of cesium-137 remained after 90.6 years, what was the mass of the original sample?

Question

10271 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-18T18:22:02

#100g#

half-life of caesium: #30.2# years

#90.6 = 30.2 * 3#

remaining mass after #90.6# years = #12.5g#
remaining mass after #3# half-lives = #12.5g#

1 half-life = original mass #* 0.5^1#

3 half-lives = original mass #* 0.5^3#

original mass #* 0.125# = #12.5g#

divide by #0.125#:

original mass = #(12.5/0.125)g#

#=12500/125g#

#=100g#

Answer 2 · 2016-12-18T18:31:47

#100 * g * Cs#

I like to use this equation:

#M(t)=M_0*(1/2)^(t/h)#

Where the mass of the substance that remains is represented as a function of t, the time which has elapsed in years, and:

#M_0# is the initial amount of the substance in gram

#t# is the amount of time which has elapsed

#h# is the half life of the substance

And in our case (we are solving for the initial amount, #M_0#:

#12.5=M_0*(1/2)^(90.6/30.2)#

So

#M_0=12.5/((1/2)^(90.6/30.2))#

and

#M_0=12.5/0.125#

#M_0=100#