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A radioactive substance decays to (1/16)th of its initial activity in 80 days. The half life of the radioactive substance expressed in days is _____.
Correct answer is '20'. Can you explain this answer?
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A radioactive substance decays to (1/16)thof its initial activity in 8...
A radioactive substance decays to (1/16)th of its initial activity in 80 days.
If N0 is the number of atoms present initially, then in 80 days, the substance decays to (1/16)th of N0 as follows:
4 × t1/2 = 80
t1/2 = 20 days
If N0 is the number of atoms present initially, then in 80 days, the substance decays to (1/16)th of N0 as follows:
4 × t1/2 = 80
t1/2 = 20 days
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A radioactive substance decays to (1/16)thof its initial activity in 8...
Given Information:
- A radioactive substance decays to (1/16)th of its initial activity in 80 days.
To Find:
- The half-life of the radioactive substance expressed in days.
Explanation:
Definition of Half-Life:
- The half-life of a radioactive substance is the time it takes for half of the radioactive atoms to decay or for the activity of the substance to decrease by half.
Using the Given Information:
- We are given that the substance decays to (1/16)th of its initial activity in 80 days.
- This means that the activity of the substance decreases by a factor of (1/16) in 80 days.
Calculating the Decay Factor:
- Let the initial activity of the substance be 'A'.
- After 80 days, the activity of the substance is (1/16) * A.
- The decay factor is given by: (1/16) * A / A = 1/16.
Calculating the Half-Life:
- The half-life can be calculated using the decay factor.
- The decay factor is related to the half-life by the equation: decay factor = (1/2)^(t/h), where 't' is the time and 'h' is the half-life.
- Substituting the given decay factor (1/16) and solving for 'h', we get: (1/16) = (1/2)^(80/h).
- Taking the logarithm of both sides, we get: log(1/16) = log((1/2)^(80/h)).
- Using the logarithmic property, we can bring the exponent down: log(1/16) = (80/h) * log(1/2).
- Simplifying, we get: -4 = (80/h) * (-1).
- Dividing both sides by -1, we get: 4 = 80/h.
- Solving for 'h', we get: h = 80/4 = 20.
Answer:
- The half-life of the radioactive substance is 20 days.
- A radioactive substance decays to (1/16)th of its initial activity in 80 days.
To Find:
- The half-life of the radioactive substance expressed in days.
Explanation:
Definition of Half-Life:
- The half-life of a radioactive substance is the time it takes for half of the radioactive atoms to decay or for the activity of the substance to decrease by half.
Using the Given Information:
- We are given that the substance decays to (1/16)th of its initial activity in 80 days.
- This means that the activity of the substance decreases by a factor of (1/16) in 80 days.
Calculating the Decay Factor:
- Let the initial activity of the substance be 'A'.
- After 80 days, the activity of the substance is (1/16) * A.
- The decay factor is given by: (1/16) * A / A = 1/16.
Calculating the Half-Life:
- The half-life can be calculated using the decay factor.
- The decay factor is related to the half-life by the equation: decay factor = (1/2)^(t/h), where 't' is the time and 'h' is the half-life.
- Substituting the given decay factor (1/16) and solving for 'h', we get: (1/16) = (1/2)^(80/h).
- Taking the logarithm of both sides, we get: log(1/16) = log((1/2)^(80/h)).
- Using the logarithmic property, we can bring the exponent down: log(1/16) = (80/h) * log(1/2).
- Simplifying, we get: -4 = (80/h) * (-1).
- Dividing both sides by -1, we get: 4 = 80/h.
- Solving for 'h', we get: h = 80/4 = 20.
Answer:
- The half-life of the radioactive substance is 20 days.
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A radioactive substance decays to (1/16)thof its initial activity in 80 days. The half life of the radioactive substance expressed in days is _____.Correct answer is '20'. Can you explain this answer?
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