The correct answer is C because:

Let D be the number of days for the number of radioactive atoms in both sources to be equal
After 5 days, the number of atoms left in X will be
- (7.3 × 10¹¹) × ${(\frac{1}{2})}^{\frac{D}{5}}$
After 6 days, the number of atoms left in Y will be
- (5.2× 10¹⁰) × ${(\frac{1}{2})}^{\frac{D}{6}}$
Equating both equations gives
- (7.3 × 10¹¹) × ${(\frac{1}{2})}^{\frac{D}{5}}$ = (5.2× 10¹⁰) × ${(\frac{1}{2})}^{\frac{D}{6}}$
Rearranging for D gives
- $\frac{7.3 \times 10^{11}}{5.2 \times 10^{10}}$ = $\frac{{(\frac{1}{2})}^{\frac{D}{6}}}{{(\frac{1}{2})}^{\frac{D}{5}}}$
- 14 = ${(\frac{1}{2})}^{\frac{D}{6} - \frac{D}{5}}$ = ${(\frac{1}{2})}^{- \frac{D}{30}}$
- ln(14) = ln $({(\frac{1}{2})}^{- \frac{D}{30}})$
- ln(14) = ${(l n (1) - l n (2))}^{- \frac{D}{30}}$ where ln(1) = 0
- ln(14) = – ${(l n (2))}^{- \frac{D}{30}}$ = $\frac{D}{30}$ ln(2)
- D = $\frac{l n (14) \times 30}{l n (2)}$ = 114 days

The key piece of logic is the ${(\frac{1}{2})}^{n u m b e r o f h a l f - l i v e s}$ . However, in this question we need to figure out the number of half–lives in this questions from the ratio $\frac{n u m b e r o f d a y s}{d a y s f o r 1 h a l f - l i f e}$ so it becomes ${(\frac{1}{2})}^{\frac{N u m b e r o f d a y s}{H a l f - l i f e}}$ . For example, if the half–life is 4 days, after 20 days, this would equal $\frac{20}{4}$ = 5 half–lives.

There are also a few key mathematical steps that have been taken in this rearranging:

1. Dividing by two powers:

When dividing by two powers $\frac{x^{a}}{x^{b}}$ = $x^{a - b}$

2. When two numbers are divided in a log, the denominator is subtracted by the numerator

log $(\frac{a}{b})$ = log(a) – log(b)

3. When a log is to a power it can also be multiplied

log(a)^b = b log(a)

Radioactive Decay (AQA A Level Physics)

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