- Joined
- Mar 23, 2003

- Messages
- 2,247

- Reaction score
- 3

(change of n / change of t) = - (decay constant)

n = n0e^-(decay constant x t)

Thanks.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter busupshot83
- Start date

- Joined
- Mar 23, 2003

- Messages
- 2,247

- Reaction score
- 3

(change of n / change of t) = - (decay constant)

n = n0e^-(decay constant x t)

Thanks.

- Joined
- Oct 25, 2006

- Messages
- 1,801

- Reaction score
- 7

It's a differential equation.

(change of n / change of t) = - (decay constant)

n = n0e^-(decay constant x t)

Thanks.

I always used 'y' instead of n so I'm going to do that for this post. Also the decay constant is generally 'k'.

Let 'change of y' = dy.

Let 'change of t' = dt.

Let 'decay constant' = k.

We have dy/dt = -ky (change of n/change of t = -decay constant * n)

This means that the rate of change of a population 'y' over a period of time 't' is proportional to the current population (ky). The negative sign means the rate is decreasing. This makes sense: over time the population will decrease.

To solve it you have to do some quick math:

dy = -ky * dt (multiply both sides by dt)

dy/y = -k * dt (divide both sides by y)

Now this can be integrated. The left side is ln and the right side is -kt. Don't forget the constants:

ln - ln(y_0) = -kt - 0 (y_0 is the initial population and the right hand side has a 0 because you integrate from 0 to t).

ln = ln(y_0) - kt

y = e^(ln(y_0) - kt)

y = e^ln(y_0) * e^(-kt)

y = y_0 * e^(-kt)

That's the equation you had. What does it mean?

Your current population is equal to the initial population times some exponential function of time. As t increases, e^(-kt) will decrease and so the population will decrease. Since t starts at 0 and e^0 = 1, the population equals y_0 initially.

How do you relate this to your original equation?

Remember you had dy/dt = -ky. So what is dy/dt? You can differentiate your function!

dy/dt (y_0 * e^(-kt)) = y_0 * (-k) * e^(-kt)

Look at what you get: -k * (y_0 e^(-kt))

Remember that y = y_0 e^(-kt), so you really end up with -ky!!

Thus, dy/dt = -ky.

How do you use the function y = y_0 e^(-kt) ?

You need to know at least 2 values. Generally they tell you that at time t = 0 the population is such-and-such. If you plug in t = 0 you get y = y_0 so that population is the initial population. Then they might tell you that at time t = 3 (for example) the population is such-and-such. So you plug in that population, you know the time, and you know y_0. And you can get k. Finally, they ask you what the population is at time t = 5 (for example). Now that you know t, k, and y_0, you can solve for y.

- Joined
- Mar 23, 2003

- Messages
- 2,247

- Reaction score
- 3

It's a differential equation.

I always used 'y' instead of n so I'm going to do that for this post. Also the decay constant is generally 'k'.

Let 'change of y' = dy.

Let 'change of t' = dt.

Let 'decay constant' = k.

We have dy/dt = -ky (change of n/change of t = -decay constant * n)

This means that the rate of change of a population 'y' over a period of time 't' is proportional to the current population (ky). The negative sign means the rate is decreasing. This makes sense: over time the population will decrease.

To solve it you have to do some quick math:

dy = -ky * dt (multiply both sides by dt)

dy/y = -k * dt (divide both sides by y)

Now this can be integrated. The left side is ln and the right side is -kt. Don't forget the constants:

ln - ln(y_0) = -kt - 0 (y_0 is the initial population and the right hand side has a 0 because you integrate from 0 to t).

ln = ln(y_0) - kt

y = e^(ln(y_0) - kt)

y = e^ln(y_0) * e^(-kt)

y = y_0 * e^(-kt)

That's the equation you had. What does it mean?

Your current population is equal to the initial population times some exponential function of time. As t increases, e^(-kt) will decrease and so the population will decrease. Since t starts at 0 and e^0 = 1, the population equals y_0 initially.

How do you relate this to your original equation?

Remember you had dy/dt = -ky. So what is dy/dt? You can differentiate your function!

dy/dt (y_0 * e^(-kt)) = y_0 * (-k) * e^(-kt)

Look at what you get: -k * (y_0 e^(-kt))

Remember that y = y_0 e^(-kt), so you really end up with -ky!!

Thus, dy/dt = -ky.

How do you use the function y = y_0 e^(-kt) ?

You need to know at least 2 values. Generally they tell you that at time t = 0 the population is such-and-such. If you plug in t = 0 you get y = y_0 so that population is the initial population. Then they might tell you that at time t = 3 (for example) the population is such-and-such. So you plug in that population, you know the time, and you know y_0. And you can get k. Finally, they ask you what the population is at time t = 5 (for example). Now that you know t, k, and y_0, you can solve for y.

I am very grateful for you taking the time out of your schedule to explain that to me. Thank you.

While your explanations helped me understand the overall concepts of the formulas

-bus

- Joined
- Oct 25, 2006

- Messages
- 1,801

- Reaction score
- 7

[226 | 88]Ra, a common isotope of radium, has a half-life of 1620 years. Knowing this, calculate the first order rate constant for the decay of radium-226 and the fraction of a sample of this isotope remaining after 100 years.

For Half-Life problems you want y = y_0 * (1/2)^(t/t_HL), where t = time elapsed and t_HL = the Half-Life of the element. You have to convert it into an exponential. You do this by noting that e^(ln x) = x. So e^ln(1/2^(t/t_HL)) = 1/2^(t/t_HL). But e^ln(1/2^(t/t_HL)) = e^((t/t_HL) * ln(1/2)) = e^((t/t_HL) * ln(2^-1)) = e^(-t/t_HL * ln 2). So k = (ln 2)/(t_HL) in this case.

Note that ln 2 = 0.693.

Fraction remaining = (y / y_0) * 100%

Answer (highlight to read): k = 0.693/1620 years = 4.28 x 10^-4/year. 95.8% of the isotope remains.

Carbon-14 is a radioactive isotope of carbon that has a half life of 5600 years. It is used extensively in dating organic material that is tens of thousands of years old. What fraction of the original amount of Carbon-14 in a sample would be present after 10,000 years?

Answer: 30%

A population of bacteria is decreasing at a rate proportional to its size. At time t = 0 hours the population consists of 1250 colonies. At time t = 5 hours the colony count is down to 700. At what time will there be 200 colonies left?

Answer: 15.8 hours

A trickier one:

A second bacterial plate contains a population of bacteria decreasing at a rate proportional to its size. At time t = 6 minutes there are 254 colonies. At time t = 8 minutes there are 242 colonies. Approximately how many colonies will there be at time t = 16 minutes? How many colonies were originally on the plate?

Answer: 200 colonies after 16 minutes. There were originally 294 colonies.

- Joined
- Mar 23, 2003

- Messages
- 2,247

- Reaction score
- 3

[226 | 88]Ra, a common isotope of radium, has a half-life of 1620 years. Knowing this, calculate the first order rate constant for the decay of radium-226 and the fraction of a sample of this isotope remaining after 100 years.

For Half-Life problems you want y = y_0 * (1/2)^(t/t_HL), where t = time elapsed and t_HL = the Half-Life of the element. You have to convert it into an exponential. You do this by noting that e^(ln x) = x. So e^ln(1/2^(t/t_HL)) = 1/2^(t/t_HL). But e^ln(1/2^(t/t_HL)) = e^((t/t_HL) * ln(1/2)) = e^((t/t_HL) * ln(2^-1)) = e^(-t/t_HL * ln 2). So k = (ln 2)/(t_HL) in this case.

Note that ln 2 = 0.693.

Fraction remaining = (y / y_0) * 100%

Answer (highlight to read): k = 0.693/1620 years = 4.28 x 10^-4/year. 95.8% of the isotope remains.

Carbon-14 is a radioactive isotope of carbon that has a half life of 5600 years. It is used extensively in dating organic material that is tens of thousands of years old. What fraction of the original amount of Carbon-14 in a sample would be present after 10,000 years?

Answer: 30%

A population of bacteria is decreasing at a rate proportional to its size. At time t = 0 hours the population consists of 1250 colonies. At time t = 5 hours the colony count is down to 700. At what time will there be 200 colonies left?

Answer: 15.8 hours

A trickier one:

A second bacterial plate contains a population of bacteria decreasing at a rate proportional to its size. At time t = 6 minutes there are 254 colonies. At time t = 8 minutes there are 242 colonies. Approximately how many colonies will there be at time t = 16 minutes? How many colonies were originally on the plate?

Answer: 200 colonies after 16 minutes. There were originally 294 colonies.

I owe you one buddy, thank you. The fact that you can explain this stuff so clear, and give such in-depth examples, reinforces the fact that you definitely know your chem . What's your background academically?

- Replies
- 2

- Views
- 917