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a radioactive source has a half life of 80s.
how long will it take for 7/8 of the source to decay?​

Respuesta :

Lets make the original number of nuclides at the start is 100.

If 7/8 of 100 is decayed, that means 87.5 decayed.

[tex] \frac{7}{8} \times 100 = 87.5[/tex]

And there is 1/8 left of the number of nuclide 100. Which is 12.5

[tex]100 - 87.5 =12.5 [/tex]

[tex] \frac{1}{8} \times 100 = 12.5[/tex]

How many Half lifes passed for 100 to become 12.5 is 3 Half-Lives.

[tex]100 \div 2 \div 2 \div 2 = 12.5[/tex]

Each Half-Life is 80 seconds so there is 240 seconds

[tex]3 \times 80 = 240[/tex]The answer is that it takes 240 seconds.
Eduard22sly

It will take 240 s.

To solve this question, we'll assume the original amount of the radioactive source to be 1.

Next, we shall determine the number of half-lives that has elapsed when 7/8 of the radioactive source has decayed. This can be obtained as illustrated below:

Original amount (N₀) = 1

Amount remaining (N) = 1 – 7/8 = 1/8

Number of half-lives (n) =?

[tex]N = \frac{N_{0} }{2^{n}} \\\\ \frac{1}{8} = \frac{1}{2^{n}}[/tex]

Cross multiply

[tex]2^{n} = 8[/tex]

Express 8 in index form with 2 as the base

[tex]2^{n} = 2^{3}[/tex]

n = 3

Thus, 3 half-lives has elapsed.

Finally, we shall determine the time taken for 7/8 of the source to decay. This can be obtained as illustrated below:

Half-life (t½) = 80 s

Number of half-lives (n) = 3

Time (t) =?

[tex]n = \frac{t}{t_{1/2}} \\ \\ 3 = \frac{t}{80}}[/tex]

Cross multiply

[tex]t = 3 * 80\\[/tex]

t = 240 s

Therefore, it will take 240 s for 7/8 of the radioactive source to decay.

R

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