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3 years ago

Thehalf life of carbon is 5730 years. how many years would it take a 200 gram sample to Decay to only 25 G?

Chemistry

1 answer:

MatroZZZ [7]3 years ago

8 0

When a half-life passes, only half of the mass of the sample stays. So to get to 25 g from 200, let's see how much half-lives will pass.

200÷2=100

100÷2=50

50÷2=25

The answer to that is 3 half-lives, or

3 × 5730 = 17190 years

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The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be

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Explanation:

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Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

Half life of the cesium-137 = $t_{1/2}=30 years[p/tex]where,
[tex]N_o$ = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the isotope

$\lambda$ = rate constant

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$N=180mg\times e^{-\frac{0.693}{30 years}\times t}$

Mass that remains after t years.

$N=180 mg\times e^{0.0231 year^{-1}\times t}$

Therefore, the parent isotope remain after one half life will be, 100 grams.

t = 70 years

$N_o=180 mg$

$t_{1/2}= 30 yeras$

$N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}$

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

$N_o=180 mg$

$t_{1/2}= 30 yeras$

N = 1 mg

t = ?

$1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}$

$\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t$

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

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Thehalf life of carbon is 5730 years. how many years would it take a 200 gram sample to Decay to only 25 G?​

Thehalf life of carbon is 5730 years. how many years would it take a 200 gram sample to Decay to only 25 G?