Question

The half-life of radioactive isotope Zn$^{65}$ is 245 days. Find the time after which activity of Zn sample remains 75% of its initial value.

Hint:

When percentage of activity is given, always convert it into a fraction before applying decay equations.

Answer 1

Correct Answer: 102

Step 1: Write the radioactive decay law

Activity A at time t is given by:

A = A₀ e^−λt

where
A₀ = initial activity
λ = decay constant

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Step 2: Use the given condition

After time t, activity remains 75% of initial value:

A = 0.75 A₀

So,

0.75 = e^−λt

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Step 3: Find the decay constant

Half-life T_1/2 = 245 days

λ = ln 2 / T_1/2

λ = 0.693 / 245

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Step 4: Solve for time t

0.75 = e^−λt

Taking natural logarithm:

ln(0.75) = −λt

t = − ln(0.75) / λ

t = − ln(0.75) × 245 / 0.693

t = (0.2877 × 245) / 0.693

t ≈ 101.6 days

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Final Answer:

The activity of the Zn sample remains 75% of its initial value after approximately
102 days