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A radioactive isotope of potassium (k) has a half-life of 20 minutes. if a 43 gram sample of this isotope is allowed to decay for 80 minutes, how many grams of the radioactive isotope will remain?

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Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 43 g

m (final mass after time T) = ? (in g)

x (number of periods elapsed) = ?

P (Half-life) = 20 minutes

T (Elapsed time for sample reduction) = 80 minutes

Let's find the number of periods elapsed (x), let us see:


T = x*P


80 = x*20


80 = 20\:x


20\:x = 80


x = (80)/(20)


\boxed{x = 4}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:


m = (m_o)/(2^x)


m = (43)/(2^(4))


m = (43)/(16)


\boxed{\boxed{m = 2.6875\:g}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

User Jack Z
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