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How much quantity of a substance is left undecayed after 19 days, if a 10. 38 g of this substance is taken initially and the half-life of this substance is 3.8 days?

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0.15 g

0.32 g

0.51 g

0.16 g

answer is B.

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Detailed Solution

$t_{1 / 2} = 3.8 days$

N₀ = 10.38 g

T = 19 days

$n = \frac{T}{t_{1} / 2} = \frac{19}{3.8} = 5$

N(remained undecayed)

$= \frac{N_{0} (initial undecayed)}{2^{n}}$

$= \frac{10.38}{2^{5}} = .32 g$

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