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guajiro [1.7K]
3 years ago
8

Phosphorus 32 has a half life of 14.3 days how much of a 4mg sample of phosphorus - 32 will remain after 71.5 days

Physics
1 answer:
andrew-mc [135]3 years ago
6 0

Let us first find out the radioactive constant of Phosphorus 32.

Radioactive constant, λ = \frac{ln2}{t_{1/2}}

Here, {t_{1/2} is half life of phosphorus 32 = 14.3 days

λ = \frac{ln2}{14.3}

The amount of 4 mg of phosphorus remain after 71.5 days can be found using the formula,

m=m₀e⁻(λt)

= 4*e^{-\frac{ln2}{14.3}71.5}

= 4*e^{-ln2*5}

=4*e^{-ln(2)^{5}}

=4*e^{-ln32}

=\frac{4}{32}

= 0.125 mg

The mass of 4 mg of phosphorus 32 remains after 71.5 days will be 0.125 mg.

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