If, initially, there is 20 mg of gold-198 (198Au) (half-life = 2.696 days),
how many days does it take for less than 1 mg to
remain?
20(.5)^(t/2.696) < 1, where t is number of days
(.5)^(t/2.696) < .05
take logs of both sides and use log rules
(t/2.696) log .5 = log .05
t/2.696 < log.05/log.5 = 4.3219..
t < 11.65 days
So it takes just less than 12 days
To determine how many days it takes for less than 1 mg of gold-198 (198Au) to remain, we can use the concept of half-life.
The half-life of gold-198 is given as 2.696 days, which means that after 2.696 days, half of the initial amount will remain.
Let's calculate how many half-lives it takes for the amount of gold-198 to decrease to less than 1 mg.
First, we divide the initial amount (20 mg) by 2 to find the amount remaining after one half-life:
20 mg / 2 = 10 mg
After the first half-life, 10 mg of gold-198 remains. Now, we repeat the process to find the amount remaining after the second half-life:
10 mg / 2 = 5 mg
After the second half-life, 5 mg of gold-198 remains. Again, we divide by 2 to find the amount remaining after the third half-life:
5 mg / 2 = 2.5 mg
After the third half-life, 2.5 mg of gold-198 remains. Once more, we divide by 2 to find the amount remaining after the fourth half-life:
2.5 mg / 2 = 1.25 mg
After the fourth half-life, 1.25 mg of gold-198 remains. Finally, we divide by 2 again to find the amount remaining after the fifth half-life:
1.25 mg / 2 = 0.625 mg
After the fifth half-life, 0.625 mg of gold-198 remains.
Since we are looking for less than 1 mg of gold-198 to remain, it takes more than five half-lives. Therefore, we can conclude that it takes more than 2.696 * 5 = 13.48 days for less than 1 mg of gold-198 to remain.
To determine how many days it takes for less than 1 mg of gold-198 to remain, we need to use the half-life of gold-198 and perform calculations to track its decay over time.
The half-life of gold-198 is given as 2.696 days. This means that every 2.696 days, half of the initial amount of gold-198 will decay.
Let's break down the problem step by step:
1. Start with the initial amount of gold-198, which is 20 mg.
2. Calculate the number of half-lives required for the amount to decrease to less than 1 mg.
To do this, we'll divide the initial amount by 2 repeatedly until we reach a value less than 1 mg.
Keep in mind that we need to convert the result of each division to mg and round it to an appropriate number of decimal places.
Here's the calculation:
First half-life: 20 mg / 2 = 10 mg
Second half-life: 10 mg / 2 = 5 mg
Third half-life: 5 mg / 2 = 2.5 mg
Fourth half-life: 2.5 mg / 2 = 1.25 mg
After the fourth half-life, the amount of gold-198 is 1.25 mg.
Based on this calculation, it took four half-lives, which is equivalent to four times the half-life of gold-198, for the amount to decrease to less than 1 mg.
To find the number of days it took, simply multiply the half-life by the number of half-lives:
Number of days = Half-life * Number of half-lives
Number of days = 2.696 days * 4
Number of days = 10.784 days
Therefore, it takes approximately 10.784 days for less than 1 mg of gold-198 to remain.