Question

Type the correct answer in the box. Express your answer to three significant figures.

Answer 1

Answer:

A tree branch that originally had 4.3 grams of carbon-14 will have  9.4586 x 10⁻³ grams after 50,000 years.

Explanation:

• The decay of carbon-14 obeys first order reaction.

The integral rate law of a first order reaction:

kt = ln ([A₀]/[A]),

where, k is the rate constant of the reaction,

t is the time of the reaction (t = 50,000 years),

[A₀] is the initial concentration of carbon-14 ([A₀] = 4.30 g).

[A] is the remaining concentration of carbon-14.

• We have for first order reactions a relation between k and half-life time (t1/2):

k = ln 2/(t1/2) = 0.693 / (5,730 years) = 1.21 x 10⁻⁴ years⁻¹.

∵ kt = ln ([A₀]/[A]).

(1.21 x 10⁻⁴ years⁻¹)(50,000 years) = ln ((4.30 g)/[A]).

6.047 = ln ((4.30 g)/[A]).

Taking the exponential of both sides:

422.893 = ((4.30 g)/[A])

∴ [A] = (4.30 g)/(422.893) = 9.4586 x 10⁻³ g.

Answer 2

Answer: The amount of C-14 isotope left after 50,000 years is 0.0101 grams.

Explanation:

All the radioactive reactions follow first order kinetics.

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

We are given:

$t_{1/2}=5730yrs$

Putting values in above equation, we get:

$k=\frac{0.693}{5730}=1.21\times 10^{-4}yr^{-1}$

The equation used to calculate amount left follows:

$N=N_o\times e^{-k\times t}$

where,

$N_o$ = initial mass of C-14 isotope = 4.3 g

N = mass of the C-14 isotope left after the time = ? g

t = time period = 50,000 years

k = rate constant = $1.21\times 10^{-4}yr^{-1}$

Putting values in above equation, we get:

$N=4.3\times e^{-(1.21\times 10^{-4}yr^{-1})\times 50,000}\\\\N=0.0101g$

Hence, the amount of C-14 isotope left after 50,000 years is 0.0101 grams.