Mixing
using the exponential distribution to calculate half-life
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Given a rock that contains 10^20 atoms of a particular substance, each atom has an exponentially distributed lifetime with a half-life of one century. How many centuries must pass before:
a) It is most likely 100 atoms remain
b) There is a 50% chance one atom remains
A solution I came across for a)
$ lambda = frac{ln(2)}{0.5} = 1.3862$
$ P(T>t) = e^{- 1.3862t} = frac{100}{10^{20}} $
$ t= 29.89 $
If $P(T>t)$ means the probability of the thing surviving until past time t, how does the expression used to calculate $P(T>t)$ equal the proportion of atoms remaining to the original amount of atoms?
Also, how would I approach problem b)?
Where would the 50% fit in for one atom remaining?
probability probability-distributions
asked Nov 10 '15 at 17:44
user288859user288859
61
$endgroup$
add a comment |
$begingroup$
Given a rock that contains 10^20 atoms of a particular substance, each atom has an exponentially distributed lifetime with a half-life of one century. How many centuries must pass before:
a) It is most likely 100 atoms remain
b) There is a 50% chance one atom remains
A solution I came across for a)
$ lambda = frac{ln(2)}{0.5} = 1.3862$
$ P(T>t) = e^{- 1.3862t} = frac{100}{10^{20}} $
$ t= 29.89 $
If $P(T>t)$ means the probability of the thing surviving until past time t, how does the expression used to calculate $P(T>t)$ equal the proportion of atoms remaining to the original amount of atoms?
Also, how would I approach problem b)?
Where would the 50% fit in for one atom remaining?
probability probability-distributions
asked Nov 10 '15 at 17:44
user288859user288859
61
$endgroup$
$begingroup$
In my opinion the value for $lambda$ is not right. The equation should be $0.5=e^{-lambda cdot 1}Rightarrow lambda=ln(2)$
$endgroup$
– callculus
Nov 10 '15 at 18:21
add a comment |
$begingroup$
Given a rock that contains 10^20 atoms of a particular substance, each atom has an exponentially distributed lifetime with a half-life of one century. How many centuries must pass before:
a) It is most likely 100 atoms remain
b) There is a 50% chance one atom remains
A solution I came across for a)
$ lambda = frac{ln(2)}{0.5} = 1.3862$
$ P(T>t) = e^{- 1.3862t} = frac{100}{10^{20}} $
$ t= 29.89 $
If $P(T>t)$ means the probability of the thing surviving until past time t, how does the expression used to calculate $P(T>t)$ equal the proportion of atoms remaining to the original amount of atoms?
Also, how would I approach problem b)?
Where would the 50% fit in for one atom remaining?
probability probability-distributions
asked Nov 10 '15 at 17:44
user288859user288859
61
$endgroup$
Given a rock that contains 10^20 atoms of a particular substance, each atom has an exponentially distributed lifetime with a half-life of one century. How many centuries must pass before:
a) It is most likely 100 atoms remain
b) There is a 50% chance one atom remains
A solution I came across for a)
$ lambda = frac{ln(2)}{0.5} = 1.3862$
$ P(T>t) = e^{- 1.3862t} = frac{100}{10^{20}} $
$ t= 29.89 $
If $P(T>t)$ means the probability of the thing surviving until past time t, how does the expression used to calculate $P(T>t)$ equal the proportion of atoms remaining to the original amount of atoms?
Also, how would I approach problem b)?
Where would the 50% fit in for one atom remaining?
probability probability-distributions
probability probability-distributions
asked Nov 10 '15 at 17:44
user288859user288859
61
asked Nov 10 '15 at 17:44
user288859user288859
61
asked Nov 10 '15 at 17:44
user288859user288859
61
asked Nov 10 '15 at 17:44
user288859user288859
61
asked Nov 10 '15 at 17:44
user288859user288859
61
61
$begingroup$
In my opinion the value for $lambda$ is not right. The equation should be $0.5=e^{-lambda cdot 1}Rightarrow lambda=ln(2)$
$endgroup$
– callculus
Nov 10 '15 at 18:21
add a comment |
$begingroup$
In my opinion the value for $lambda$ is not right. The equation should be $0.5=e^{-lambda cdot 1}Rightarrow lambda=ln(2)$
$endgroup$
– callculus
Nov 10 '15 at 18:21
$begingroup$
In my opinion the value for $lambda$ is not right. The equation should be $0.5=e^{-lambda cdot 1}Rightarrow lambda=ln(2)$
$endgroup$
– callculus
Nov 10 '15 at 18:21
$begingroup$
In my opinion the value for $lambda$ is not right. The equation should be $0.5=e^{-lambda cdot 1}Rightarrow lambda=ln(2)$
$endgroup$
– callculus
Nov 10 '15 at 18:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As stated in the comments, there is an error in $lambda$ (unless the halflife is supposed to be half a century and you made a typo).
"Exponentially distributed lifetime" means that the number of atoms in the rock follows an evolution of the form
$$
N(t) = N_0 e^{-lambda t},
$$
where $N$ is the number of atoms at time $t$ and $N_0=N(0)$.
To find $lambda$, recall that after one halflife ($t=1$ centuries) half of the initial number of atoms remain, thus
$$
frac{N_0}{2} = N_0 e^{-lambda} Leftrightarrow lambda = log{(2)}.
$$
(a) Put $N(t)=100$, $N_0 = 10^{20}$, $lambda = log{(2)}$ and solve for $t$. You will obtain the same equation as you have written in the question.
(b) Recall that a halflife is the time during which the probability of decay of one atom is 50%. Thus, set $N(t)=1/2$, $N_0=10^{20}$, $lambda={(2)}$ and solve for $t$.
answered Nov 11 '15 at 10:14
ekkilopekkilop
1,736519
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
As stated in the comments, there is an error in $lambda$ (unless the halflife is supposed to be half a century and you made a typo).
"Exponentially distributed lifetime" means that the number of atoms in the rock follows an evolution of the form
$$
N(t) = N_0 e^{-lambda t},
$$
where $N$ is the number of atoms at time $t$ and $N_0=N(0)$.
To find $lambda$, recall that after one halflife ($t=1$ centuries) half of the initial number of atoms remain, thus
$$
frac{N_0}{2} = N_0 e^{-lambda} Leftrightarrow lambda = log{(2)}.
$$
(a) Put $N(t)=100$, $N_0 = 10^{20}$, $lambda = log{(2)}$ and solve for $t$. You will obtain the same equation as you have written in the question.
(b) Recall that a halflife is the time during which the probability of decay of one atom is 50%. Thus, set $N(t)=1/2$, $N_0=10^{20}$, $lambda={(2)}$ and solve for $t$.
answered Nov 11 '15 at 10:14
ekkilopekkilop
1,736519
$endgroup$
add a comment |
$begingroup$
As stated in the comments, there is an error in $lambda$ (unless the halflife is supposed to be half a century and you made a typo).
"Exponentially distributed lifetime" means that the number of atoms in the rock follows an evolution of the form
$$
N(t) = N_0 e^{-lambda t},
$$
where $N$ is the number of atoms at time $t$ and $N_0=N(0)$.
To find $lambda$, recall that after one halflife ($t=1$ centuries) half of the initial number of atoms remain, thus
$$
frac{N_0}{2} = N_0 e^{-lambda} Leftrightarrow lambda = log{(2)}.
$$
(a) Put $N(t)=100$, $N_0 = 10^{20}$, $lambda = log{(2)}$ and solve for $t$. You will obtain the same equation as you have written in the question.
(b) Recall that a halflife is the time during which the probability of decay of one atom is 50%. Thus, set $N(t)=1/2$, $N_0=10^{20}$, $lambda={(2)}$ and solve for $t$.
answered Nov 11 '15 at 10:14
ekkilopekkilop
1,736519
$endgroup$
add a comment |
$begingroup$
As stated in the comments, there is an error in $lambda$ (unless the halflife is supposed to be half a century and you made a typo).
"Exponentially distributed lifetime" means that the number of atoms in the rock follows an evolution of the form
$$
N(t) = N_0 e^{-lambda t},
$$
where $N$ is the number of atoms at time $t$ and $N_0=N(0)$.
To find $lambda$, recall that after one halflife ($t=1$ centuries) half of the initial number of atoms remain, thus
$$
frac{N_0}{2} = N_0 e^{-lambda} Leftrightarrow lambda = log{(2)}.
$$
(a) Put $N(t)=100$, $N_0 = 10^{20}$, $lambda = log{(2)}$ and solve for $t$. You will obtain the same equation as you have written in the question.
(b) Recall that a halflife is the time during which the probability of decay of one atom is 50%. Thus, set $N(t)=1/2$, $N_0=10^{20}$, $lambda={(2)}$ and solve for $t$.
answered Nov 11 '15 at 10:14
ekkilopekkilop
1,736519
$endgroup$
As stated in the comments, there is an error in $lambda$ (unless the halflife is supposed to be half a century and you made a typo).
"Exponentially distributed lifetime" means that the number of atoms in the rock follows an evolution of the form
$$
N(t) = N_0 e^{-lambda t},
$$
where $N$ is the number of atoms at time $t$ and $N_0=N(0)$.
To find $lambda$, recall that after one halflife ($t=1$ centuries) half of the initial number of atoms remain, thus
$$
frac{N_0}{2} = N_0 e^{-lambda} Leftrightarrow lambda = log{(2)}.
$$
(a) Put $N(t)=100$, $N_0 = 10^{20}$, $lambda = log{(2)}$ and solve for $t$. You will obtain the same equation as you have written in the question.
(b) Recall that a halflife is the time during which the probability of decay of one atom is 50%. Thus, set $N(t)=1/2$, $N_0=10^{20}$, $lambda={(2)}$ and solve for $t$.
answered Nov 11 '15 at 10:14
ekkilopekkilop
1,736519
answered Nov 11 '15 at 10:14
ekkilopekkilop
1,736519
answered Nov 11 '15 at 10:14
ekkilopekkilop
1,736519
answered Nov 11 '15 at 10:14
ekkilopekkilop
1,736519
1,736519
add a comment |
add a comment |
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$begingroup$
In my opinion the value for $lambda$ is not right. The equation should be $0.5=e^{-lambda cdot 1}Rightarrow lambda=ln(2)$
$endgroup$
– callculus
Nov 10 '15 at 18:21