Question

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?

Answer 1

Answer:

3.21 grams

Step-by-step explanation:

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

lambda = $\frac{ln(2)}{half-life}$

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 = $\frac{ln(2)}{13hours}$

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 $3days(\frac{24hours}{1day} )=72hours$; 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.