To solve this problem, we need to use the formula for radioactive decay which is:

N = N0 * (1/2)^(t/T)

where:
N is the final quantity of the substance
N0 is the initial quantity of the substance
t is the time that has passed
T is the half-life of the substance

Given:
N0 = 2.20 mg
t = 117 years
T = 28.8 years

Substituting these values into the formula, we get:

N = 2.20 mg * (1/2)^(117/28.8)

Calculating the exponent first:

117/28.8 = 4.06

So, the equation becomes:

N = 2.20 mg * (1/2)^4.06

Calculating the second part:

(1/2)^4.06 = 0.060

So, the equation becomes:

N = 2.20 mg * 0.060

Finally, multiplying these values together gives:

N = 0.132 mg

Rounding to two significant digits, we get:

N = 0.13 mg

So, after 117 years, 0.13 mg of the Strontium-90 sample would be left.

Question

To solve this problem, we need to use the formula for radioactive decay which is:

N = N0 * (1/2)^(t/T)

where:
N is the final quantity of the substance
N0 is the initial quantity of the substance
t is the time that has passed
T is the half-life of the substance

Given:
N0 = 2.20 mg
t = 117 years
T = 28.8 years

Substituting these values into the formula, we get:

N = 2.20 mg * (1/2)^(117/28.8)

Calculating the exponent first:

117/28.8 = 4.06

So, the equation becomes:

N = 2.20 mg * (1/2)^4.06

Calculating the second part:

(1/2)^4.06 = 0.060

So, the equation becomes:

N = 2.20 mg * 0.060

Finally, multiplying these values together gives:

N = 0.132 mg

Rounding to two significant digits, we get:

N = 0.13 mg

So, after 117 years, 0.13 mg of the Strontium-90 sample would be left.

Knowee AI · Accepted Answer

To solve this problem, we need to use the formula for radioactive decay which is:

N = N0 * (1/2)^(t/T)

where:
N is the final quantity of the substance
N0 is the initial quantity of the substance
t is the time that has passed
T is the half-life of the substance

Given:
N0 = 2.20 mg
t = 117 years
T = 28.8 years

Substituting these values into the formula, we get:

N = 2.20 mg * (1/2)^(117/28.8)

Calculating the exponent first:

117/28.8 = 4.06

So, the equation becomes:

N = 2.20 mg * (1/2)^4.06

Calculating the second part:

(1/2)^4.06 = 0.060

So, the equation becomes:

N = 2.20 mg * 0.060

Finally, multiplying these values together gives:

N = 0.132 mg

Rounding to two significant digits, we get:

N = 0.13 mg

So, after 117 years, 0.13 mg of the Strontium-90 sample would be left.

Strontium-90 is radioactive and has a half life of 28.8 years. How much of a 2.20mg sample would be left after 117. years?Round your answer to 2 significant digits. Also, be sure your answer has a unit symbol.

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