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Samarium has a half-life of approximately 106 billion years. If you found a sample of samarium from the beginning of Earth's history, approximately 4.6 billion years ago, which would be the best approximation of the amount of parent isotope left in the rock?
a Nearly 100 percent
b 50 percent
c 25 percent
d Less than 10 percent
Brainliest for the correct answer and 20 pts~!

Respuesta :

AL2006

4.3 billion years is only 4.3% of the time it takes for half of the Samarium to decay, so we would expect nearly all (100%) of the parent isotope to still be present after that short (? ! ?) an amount of time.

Answer:

A) Nearly 100 percent

Explanation:

As we know that the amount of radioactive substance is given at any time by the equation

[tex]N = N_o e^{-\lambda t}[/tex]

here we know that

[tex]\lambda = \frac{ln2}{T_{1/2}}[/tex]

we know that

[tex]T_{1/2} = 106 billion\: years[/tex]

now radioactive constant is given as

[tex]T_{1/2} = \frac{ln2}{106} = 6.54 \times 10^{-3}[/tex]

now number present after t = 4.6 billion years

[tex]N = N_oe^{-6.54 \times 10^{-3}(4.6)}[/tex]

[tex]\frac{N}{N_o} = 0.97[/tex]

so it is approximately 97% amount is left after given time

so correct answer would be

Nearly 100 percent

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