Question

Radioactive gold-198 is used in the diagnosis of liver problems. 198Au decays in a first-order process, emitting a β particle (electron). The half-life of this isotope is 2.7 days. You begin with a 5.6-mg sample of the isotope. Calculate how much gold-198 remains after 3.0 days.

Answer 1

Nt = 2.59 mg

The amount of gold-198 remaining after 3.0 days is 2.59 mg

Step-by-step explanation:

Decay of a substance can be expressed mathematically as

Nt = No.e^(-λt) .....1

Where

Nt = amount remaining at time t

No = initial amount = 5.6 mg

t = decay time = 3 days

λ = decay constant

And decay constant can be expressed in terms of half life as;

λ = ln(2)/tₕ ....2

Where;

tₕ = half life = 2.7 days

Substituting equation 2 to 1.

Nt = No.e^(-t × ln(2)/tₕ)

Substituting the values, we have;

Nt = 5.6.e^(-3 × ln(2)/2.7)

Nt = 2.59 mg

The amount of gold-198 remaining after 3.0 days is 2.59 mg

Answer 2

2.576 mg of gold-198 remains after 3 days

Step-by-step explanation:

N = No(0.5)^t/t1/2

No is the initial amount of gold-198 = 5.6 mg

t is the time taken for gold-198 to reduce to a certain amount (N) = 3 days

t1/2 is the half-life of gold-198 = 2.7 days

N = 5.6(0.5)^3/2.7 = 5.6(0.5)^1.11 = 5.6×0.46 = 2.576 mg