Answer:
3.21 grams
Step-by-step explanation:
First we are finding the radioactive decay constant using the formula:
lambda = [tex]\frac{ln(2)}{half-life}[/tex]
where
lambda is the radioactive decay constant.
[tex]half-life[/tex] 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 = [tex]\frac{ln(2)}{13hours}[/tex]
lambda = 0.0533 per hour
Now we can use the decay formula to find the remaining quantity of the substance:
[tex]N_t=N_0e^{(-lamda)(t)[/tex]
where
[tex]N_t[/tex] is the ending amount
[tex]N_0[/tex] is the beginning amount
[tex]t[/tex] 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 [tex]N_0=149[/tex]. Notice that there are exactly 3 days from 12 am October 19th to 12 am October 22nd, so we have [tex]3days(\frac{24hours}{1day} )=72hours[/tex]; therefore [tex]t=72[/tex]. Now we can replace all the values in our formula:
[tex]N_t=149e^{(-0.0533)(72)[/tex]
[tex]N_t=3.21[/tex]
We can conclude that 3.21 grams of Rutherfordium- 265 would remain on October 22nd at 12 am.
