In a radioactive material the activity at time t1, is A1 and at a later time t2, it is A2. If the decay constant of the material is λ, thenA1=A2e−λ(t1−t2)A1=A2eλ(t1−t2)A1=A2(t2/t1)A1=A2
Question
In a radioactive material the activity at time t1, is A1 and at a later time t2, it is A2. If the decay constant of the material is λ, thenA1=A2e−λ(t1−t2)A1=A2eλ(t1−t2)A1=A2(t2/t1)A1=A2
Solution
The correct equation that describes the relationship between the activity of a radioactive material at two different times is A1=A2e^−λ(t2−t1).
Here's the step by step explanation:
-
The activity of a radioactive material is given by the equation A = A0e^−λt, where A is the activity at time t, A0 is the initial activity, λ is the decay constant, and t is the time.
-
If we want to find the activity at a later time t2, we can substitute t2 into the equation to get A2 = A0e^−λt2.
-
Similarly, the activity at an earlier time t1 is given by A1 = A0e^−λt1.
-
We can set these two equations equal to each other to find the relationship between A1 and A2: A1/A2 = e^−λ(t2−t1).
-
Therefore, A1 = A2e^−λ(t2−t1) is the correct equation.
Similar Questions
A rate of decay of a radioactive element is independent of temperature, pressure, Infact for any external conditions. A simple law discovered by Rutherford states that a certain fraction of any sample of radioactive element undergoes change in a unit time. If we start with 10mg of the radioactive substance (t12 = 1day) , only 5mg will be left at the end of the first day. At the end of 2nd day 2.5 mg will be left and so on. λ = 2.303tlogaa − x .The reaction , is an example of 12D + 13T → 24He + 01n is an example of
A rate of decay of a radioactive element is independent of temperature, pressure, Infact for any external conditions. A simple law discovered by Rutherford states that a certain fraction of any sample of radioactive element undergoes change in a unit time. If we start with 10 mg of the radioactive substance (t12 = 1day) , only 5mg will be left at the end of the first day. At the end of 2nd day 2.5 mg will be left and so on. λ = 2.303tlogaa − x A human body required the 0.01μ activity of radioactive substance after 24 hours. Half life of radioactive is 6 hours. Then injection of maximum activity of radioactive substance that can be injected:
radioactive isotope decays with a rate proportional to the original amountof material present. If originally there is 50mg of the material present and aftertwo hours it has lost 10% of its mass, determine the half-life, the time at which itis half its original mass.
The activity of a radioactive material is 2.56 × 10–3 Ci. If the half life of the material is 5 days, after how many days the activity will become 2 × 10–5 Ci
The activity of a radioactive source is the rate at which the nuclei of the source decay. What is the unit for the activity of a radioactive source?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.