Question

Find the half-life of an element which decays by 3.411% each day. Hint: use y = ab^t.

Answer 1

a/2=a((100-3.411)/100)^t

a/2=a(0.96589)^t

0.5=0.96589^t 

ln(0.5)=t(ln(0.96589))

t=ln(0.5)/ln(0.96589)

t≈19.97 days (to nearest hundredth of a day)

Answer 2

This is about half life of elements with exponential decay.

Half life = 20 years

• We are given a decay rate of 3.411% per day.

We are given;

y = ab^(t)

Where;

t is the half life

y = a/2 is the amount of substance remaining after decay

a is amount of substance initially

b = 100% - 3.411% = 96.589% = 0.96589

• Thus;

a/2 = a(0.96589)^(t)

a will cancel out to give;

0.5 = 0.96589^(t)

ln (0.5) = t(ln 0.96589)

t = ln(0.5)/ln(0.96589)

t = 19.968 days

This is approximately 20 days.

Read more at; brainly.com/question/13891943