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Elenna [48]
3 years ago
12

Radon, the radioactive gas that contributes more than half the yearly dose to the average American, has a half-life of approxima

tely 4 days. How many days does it take a radon source to decay to 1/128 of its original level of radioactivity?
Physics
1 answer:
Nadusha1986 [10]3 years ago
3 0

Answer:

It takes 28 days.

Explanation:

Hi there!

The half-life is the time at which the concentration of a substance is halved. For radon, it is 4 days. Then, after 4 days, the amount of radon present in the source will be 1/2 of its original concentration. After another 4 days, the concentration of radon will be half of the half of its original concentration, that is, 1/4 of the amount of radon originally present.

After another 4 days, the concentration will be halved again (1/4 /2 = 1/8) and after another 4 days, it will be halved again (1/8 / 2 = 1/16) and so on.

Then, after n half-lives, the concentration of radon will be reduced by a factor of 1/2ⁿ.

Then, to know how many half-lives it takes for the concentration of radon to be reduced to 1/128 of its original level, let´s solve this equation for n:

1/2ⁿ = 1/128

2ⁿ = 128

log 2ⁿ = log 128

n log 2 = log 128

n = log 128 / log 2

n = 7

After 7 half-lives, the concentration of radon will be reduced by 1/128. Since a half-life is 4 days, 7 half-lives will be (7 · 4) 28 days.

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