Question

Polonium-218 has a half-life of about 2 days.

After 10 days, how many milligrams of a 1200 mg sample will remain?

Answer 1

37.5 mg

Explanation:

So, The half life of plutonium = 2 days

Original amount = 1200 mg

So the number of halflive periods in 10 days = 10/2 =5

SO we use the formula:

`(P)/(2^n)`

Where P is the original amount and n is the amount of halflife periods.

By substituting we get:

`(1200)/(2^5) \\\\(1200)/(32)=37.5`

Answer 2

37.5 mg

Explanation:

Half life of Polonium-218 = 2 days

Initial sample mass = 1200 mg

Time interval = 10 days.

Number of half lives in 10 days =
`\[(10)/(2)\]` = 5

After 5 half lives the initial sample will reduce to
`\[(1)/(2^(5))\]` of its initial amount.

So 1200 mg will reduce to
`\[(1200)/(2^(5))\]` = 37.5 mg

The resultant sample weight will become 37.5 milligram after 10 days.