The half-life of cesium-137 is 30 years. Suppose we have a 120-mg sample. Find the mass that remains after #t# years. How much of the sample remains after 90 years? After how long will only 1 mg remain?

Question

82569 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-10T23:46:20

#(a) "120 mg"/2^(t/"30 yr"); (b) "15.0 mg"; (c) "199 yr"#

(a) Mass after t years

The formula for calculating the amount remaining after a number of half-lives, #n# is

#color(blue)(bar(ul(|color(white)(a/a)"A" = "A"_0/2^ncolor(white)(a/a)|)))" "#

where

#"A"_0# is the initial amount,

#"A"# is the amount remaining and

#n = t/t_½#

This gives

#"A" = A_0/2^( t/t_½)#

∴ #"A" = "120 mg"/2^(t/"30 yr")#

(b) Mass after 90 yr

#"A" = "120 mg"/2^(t/"30 yr") = "120 mg"/2^(("90 yr")/("30 yr")) = "120 mg"/2^3 = "120 mg"/8 = "15.0 mg"#

(c) Time for 1 mg remaining

#"A" = A_0/2^( t/t_½)#

#A_0/"A" = 2^( t/t_½)#

#(100 color(red)(cancel(color(black)("mg"))))/(1 color(red)(cancel(color(black)("mg")))) = 2^(t/"30 yr")#

#2^(t/"30 yr") = 100#

#(t/"30 yr")log2 = log100 =2#

#t/"30 yr" = 2/log2 =6.64#

#t = "6.64 × 30 yr" = "199 yr"#