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Rutherfordium- 265 is a radioactive substance that has a half-life of 13 hours. If there is a 149g sample of Rutherfordium- 265 left at 12am on October 19th, approximately how much Rutherfordium- 265 would remain on October 22nd at 12am?

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4 votes

Answer:

3.21 grams

Explanation:

First we are finding the radioactive decay constant using the formula:

lambda =
(ln(2))/(half-life)

where

lambda is the radioactive decay constant.


half-life is the half life of the radioactive substance.

We know from our problem that Rutherfordium- 265 has a half-life of 13 hours, so let's replace the value in our formula.

lambda =
(ln(2))/(13hours)

lambda = 0.0533 per hour

Now we can use the decay formula to find the remaining quantity of the substance:


N_t=N_0e^{(-lamda)(t)

where


N_t is the ending amount


N_0 is the beginning amount


t is the time (in hours)

We know from our problem that there is a 149g sample of Rutherfordium- 265 left at 12am on October 19th, so
N_0=149. Notice that there are exactly 3 days from 12 am October 19th to 12 am October 22nd, so we have
3days((24hours)/(1day) )=72hours; therefore
t=72. Now we can replace all the values in our formula:


N_t=149e^{(-0.0533)(72)


N_t=3.21

We can conclude that 3.21 grams of Rutherfordium- 265 would remain on October 22nd at 12 am.

User JoeBobs
by
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