Answer :
Certainly! Let's solve this problem step-by-step.
Problem Statement: A sample of radioactive material has a half-life of 12 years. How long will it take for a 50g sample to decay to 6.25g?
First, let's understand the concept of half-life. The half-life of a radioactive substance is the time required for half of the material to decay.
To solve this, we'll use the understanding of exponential decay. The formula for exponential decay in relation to half-life is:
[tex]\[ \text{final\_mass} = \text{initial\_mass} \times (0.5)^{\frac{\text{time\_elapsed}}{\text{half\_life}}} \][/tex]
Here:
- [tex]\(\text{initial\_mass} = 50 \, \text{grams}\)[/tex]
- [tex]\(\text{final\_mass} = 6.25 \, \text{grams}\)[/tex]
- [tex]\(\text{half\_life} = 12 \, \text{years}\)[/tex]
We need to find the [tex]\(\text{time\_elapsed}\)[/tex].
First, rearrange the formula to solve for [tex]\(\text{time\_elapsed}\)[/tex]:
[tex]\[ (0.5)^{\frac{\text{time\_elapsed}}{\text{half\_life}}} = \frac{\text{final\_mass}}{\text{initial\_mass}} \][/tex]
Calculate the fractional mass:
[tex]\[ \frac{6.25 \, \text{grams}}{50 \, \text{grams}} = 0.125 \][/tex]
So our equation now reads:
[tex]\[ (0.5)^{\frac{\text{time\_elapsed}}{12}} = 0.125 \][/tex]
To solve the above equation, we'll take the logarithm of both sides. The choice of the base for logarithms is typically base 10 or natural logarithm (base [tex]\(e\)[/tex]), but it doesn't affect the result as long as we are consistent.
[tex]\[ \log((0.5)^{\frac{\text{time\_elapsed}}{12}}) = \log(0.125) \][/tex]
By logarithmic properties, bring the exponent to the front:
[tex]\[ \frac{\text{time\_elapsed}}{12} \cdot \log(0.5) = \log(0.125) \][/tex]
Next, solve for [tex]\(\text{time\_elapsed}\)[/tex]:
[tex]\[ \text{time\_elapsed} = \frac{12 \cdot \log(0.125)}{\log(0.5)} \][/tex]
Using the given values and the result we know to be correct:
[tex]\[ \text{time\_elapsed} \approx 36 \, \text{years} \][/tex]
Therefore, it will take approximately 36 years for a 50g sample of the radioactive material to decay to 6.25g.
Problem Statement: A sample of radioactive material has a half-life of 12 years. How long will it take for a 50g sample to decay to 6.25g?
First, let's understand the concept of half-life. The half-life of a radioactive substance is the time required for half of the material to decay.
To solve this, we'll use the understanding of exponential decay. The formula for exponential decay in relation to half-life is:
[tex]\[ \text{final\_mass} = \text{initial\_mass} \times (0.5)^{\frac{\text{time\_elapsed}}{\text{half\_life}}} \][/tex]
Here:
- [tex]\(\text{initial\_mass} = 50 \, \text{grams}\)[/tex]
- [tex]\(\text{final\_mass} = 6.25 \, \text{grams}\)[/tex]
- [tex]\(\text{half\_life} = 12 \, \text{years}\)[/tex]
We need to find the [tex]\(\text{time\_elapsed}\)[/tex].
First, rearrange the formula to solve for [tex]\(\text{time\_elapsed}\)[/tex]:
[tex]\[ (0.5)^{\frac{\text{time\_elapsed}}{\text{half\_life}}} = \frac{\text{final\_mass}}{\text{initial\_mass}} \][/tex]
Calculate the fractional mass:
[tex]\[ \frac{6.25 \, \text{grams}}{50 \, \text{grams}} = 0.125 \][/tex]
So our equation now reads:
[tex]\[ (0.5)^{\frac{\text{time\_elapsed}}{12}} = 0.125 \][/tex]
To solve the above equation, we'll take the logarithm of both sides. The choice of the base for logarithms is typically base 10 or natural logarithm (base [tex]\(e\)[/tex]), but it doesn't affect the result as long as we are consistent.
[tex]\[ \log((0.5)^{\frac{\text{time\_elapsed}}{12}}) = \log(0.125) \][/tex]
By logarithmic properties, bring the exponent to the front:
[tex]\[ \frac{\text{time\_elapsed}}{12} \cdot \log(0.5) = \log(0.125) \][/tex]
Next, solve for [tex]\(\text{time\_elapsed}\)[/tex]:
[tex]\[ \text{time\_elapsed} = \frac{12 \cdot \log(0.125)}{\log(0.5)} \][/tex]
Using the given values and the result we know to be correct:
[tex]\[ \text{time\_elapsed} \approx 36 \, \text{years} \][/tex]
Therefore, it will take approximately 36 years for a 50g sample of the radioactive material to decay to 6.25g.
