The half-life of iodine-131 is 13 hours. If a sample of radium-226 has an original activity of 400 Bg, what will its activity be after:
i) 26 hours?
ii) 39 hours?
iii) 52 hours

I. 100 Bg

II. 50 Bg

III. 25 Bg

Explanation:

I. Determination of the activity after 26 hours.

We'll begin by calculating the number of half-lives that has elapse. This can be obtained as follow:

Half-life (t½) = 13 hours

Time (t) = 26 hours

Number of half-lives (n) =?

n = t / t½

n = 26 / 13

n = 2

Finally, we shall determine remaining activity

Original activity (N₀) = 400 Bg

Number of half-lives (n) = 2

Activity remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2² × 400

N = 1/4 × 400

N = 100 Bg

II. Determination of the activity after 39 hours.

We'll begin by calculating the number of half-lives that has elapse. This can be obtained as follow:

Half-life (t½) = 13 hours

Time (t) = 39 hours

Number of half-lives (n) =?

n = t / t½

n = 39 / 13

n = 3

Finally, we shall determine remaining activity.

Original activity (N₀) = 400 Bg

Number of half-lives (n) = 3

Activity remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2³ × 400

N = 1/8 × 400

N = 50 Bg

III. Determination of the activity after 52 hours.

We'll begin by calculating the number of half-lives that has elapse. This can be obtained as follow:

Half-life (t½) = 13 hours

Time (t) = 52 hours

Number of half-lives (n) =?

n = t / t½

n = 52 / 13

n = 4

Finally, we shall determine remaining activity

Original activity (N₀) = 400 Bg

Number of half-lives (n) = 4

Activity remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2⁴ × 400

N = 1/16 × 400

N = 25 Bg

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.

mistakes are ok