Answered

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Element X is a radioactive isotope such that every 22 years, its mass decreases by half. Given that the initial mass of a sample of Element X is 1200 grams, how long would it be until the mass of the sample reached 800 grams, to the nearest tenth of a year?

Sagot :

reinhard10158

Step-by-step explanation:

s1 = 1200 g

s2 = 1200×2^(-1/22)

s3 = 1200×2^(-2/22)

...

s23 (22 years later) = 1200×2^-22/22 ≈ 1200×2^-1 = 1200/2

sn = 1200×2^(-(n-1)/22)

what is the n, so that the result is 800 g ?

800 = 1200×2^(-(n-1)/22)

800/1200 = 2^(-(n-1)/22)

2/3 = 2^(-(n-1)/22)

log2(2/3) = -(n-1)/22

22×log2(2/3) = -(n - 1) = -n + 1

n = -22×log2(2/3) + 1 = 13.86917502...

since we stared counting with n=1 for the starting quantity, the number of years is truly 1 less than n (s2 is after 1 year, s3 after 2 years ...).

so, we know for n = 13.86917502..., that in fact

12.86917502... ≈ 12.9 years have passed until the mass of the sample reached 800g.

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