Question

An original sample of the radioisotope fluorine-21 had a mass of 80.0 milligrams. Only 20.0 milligrams of this original sample remain unchanged after 8.32 seconds. What is the half-life of fluorine-21?
(1) 1.04 s (3) 4.16 s
(2) 2.08 s (4) 8.32 s

Answer 1

Answer: The correct answer is Option 3.

Explanation:

All the radioisotope decay processes follow first order kinetics.

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,  

k = rate constant  = ?

t = time taken for decay process = 8.32 seconds

a = initial amount of the reactant  = 80 mg

a - x = amount left after decay process  = 20 mg

Putting values in above equation, we get:

$k=\frac{2.303}{8.32sec}\log\frac{80g}{20}\\\\k=0.166sec^{-1}$

The equation used to calculate half life for first order kinetics:

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

where,

$t_{1/2}$ = half life of the reaction = ?

k = $0.166sec^{-1}$

Putting values in above equation, we get:

$t_{1/2}=\frac{0.693}{0.166sec^{-1}}=4.16sec$

Hence, the correct answer is Option 3.

Answer 2

Answer:

4.16s

Explanation:

N = 20*10⁻³g

N₀ = 80*10⁻³g

t = 8.32

N = N₀e⁻λt

In(N/N₀) = -λt

-λ = 1/t * In(N/N₀)

-λ = 1 / 8.32 * In (20*10⁻³ / 80*10⁻³)

-λ = 0.12 * In(0.25)

-λ = -0.167

λ = 0.167

t½ = 0.693 / λ

t½ = 0.693 / 0.167

t½ = 4.16s