An isotope has an initial activity of 120 Bq. 6 days later its activity is 15 Bq. The half-life is?

Assessment: JAMB UTME - Physics - 2022 Subject: Physics

Question 1 Report

3 days

2 days

1 day

4 days

Answer Details

The equation that establishes a relationship between the amount left undecayed,

A, the initial amount, A0, and the number of half-lives that pass in a period of time t looks like this:

$\frac{1}{2^{n}} * A_{0} = A$

$\frac{1}{2^{n}} * 120 = 15$

$\frac{1}{2^{n}}$ = $\frac{15}{120}$

$\frac{1}{2^{n}}$ = $\frac{1}{8}$

$\frac{1}{2^{n}}$ = $\frac{1}{2^{3}}$

$2^{n} = 2^{3}$

n = 3
:6days ÷ 3 = 2days

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