A mosquito fed on a solution containing phosphorous-32 is released. Phosphorous-32 has a half-life of 14 days. When the mosquito is recaptured 28 days later, what percentage of the original Phosphorus-32 will remain?

28 days is 2 half lives.

Start P-32 = 100%
1st half life = 1/2 that or 50% left
2nd half life = 1/2 that or 25% left.

To determine the percentage of the original Phosphorus-32 that will remain after 28 days, we need to consider its half-life. The half-life of Phosphorus-32 is 14 days, meaning that after 14 days, half of the original amount will decay.

To find out how many half-lives have occurred over the 28-day period, we divide the total time by the half-life:

Number of half-lives = 28 days / 14 days = 2

Since two half-lives have passed, we know that the original amount of Phosphorus-32 has been halved twice. Each half-life reduces the amount to 50% of its previous value. So, after two half-lives, the percentage of the original Phosphorus-32 remaining is:

Percentage remaining = 50% x 50% = 25%

Therefore, after 28 days, only 25% of the original Phosphorus-32 will still be present.