Radium has a half-life of 1580 years. If a sample contains 1g of radium now, how much radium will there be:
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a) In 500 years
b) In 3000 years
c) 1000 years ago
I know that this is a half life question but am not sure what the formula is and how to solve part a, b, and c. Can somebody please show me the steps?
exponential-function
asked Nov 23 at 2:24
Andrew Nelson
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a) In 500 years
b) In 3000 years
c) 1000 years ago
I know that this is a half life question but am not sure what the formula is and how to solve part a, b, and c. Can somebody please show me the steps?
exponential-function
asked Nov 23 at 2:24
Andrew Nelson
11
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0
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favorite
up vote
0
down vote
favorite
a) In 500 years
b) In 3000 years
c) 1000 years ago
I know that this is a half life question but am not sure what the formula is and how to solve part a, b, and c. Can somebody please show me the steps?
exponential-function
asked Nov 23 at 2:24
Andrew Nelson
11
a) In 500 years
b) In 3000 years
c) 1000 years ago
I know that this is a half life question but am not sure what the formula is and how to solve part a, b, and c. Can somebody please show me the steps?
exponential-function
exponential-function
asked Nov 23 at 2:24
Andrew Nelson
11
asked Nov 23 at 2:24
Andrew Nelson
11
asked Nov 23 at 2:24
Andrew Nelson
11
asked Nov 23 at 2:24
Andrew Nelson
11
asked Nov 23 at 2:24
Andrew Nelson
11
11
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2 Answers
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Hint: Let $H$ (for half-life) be $1580$ years. Then if $f(t)$ is the remaining quantity in grams after a time $t$, we have:
$$begin{align}f(H) &= frac12\
f(2H) &= frac1{2^2}\
f(3H) &= frac1{2^3}, textrm{etc.}
end{align}$$
In general, we see that $f(xH) = 2^{-x}$. Now try to manipulate this to be able to work with the times you've been given.
answered Nov 23 at 2:33
Théophile
19.4k12946
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Let $D$ be the rate of decay. Then we have that:
$$e^{1580D}=frac 12$$
You can use this to find $D=frac{-ln(2)}{1580}$
Then you simply need to plug in $D$ for $e^{500D}$ etc
answered Nov 23 at 2:47
Rhys Hughes
4,6751327
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2 Answers
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2 Answers
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Hint: Let $H$ (for half-life) be $1580$ years. Then if $f(t)$ is the remaining quantity in grams after a time $t$, we have:
$$begin{align}f(H) &= frac12\
f(2H) &= frac1{2^2}\
f(3H) &= frac1{2^3}, textrm{etc.}
end{align}$$
In general, we see that $f(xH) = 2^{-x}$. Now try to manipulate this to be able to work with the times you've been given.
answered Nov 23 at 2:33
Théophile
19.4k12946
add a comment |
up vote
0
down vote
Hint: Let $H$ (for half-life) be $1580$ years. Then if $f(t)$ is the remaining quantity in grams after a time $t$, we have:
$$begin{align}f(H) &= frac12\
f(2H) &= frac1{2^2}\
f(3H) &= frac1{2^3}, textrm{etc.}
end{align}$$
In general, we see that $f(xH) = 2^{-x}$. Now try to manipulate this to be able to work with the times you've been given.
answered Nov 23 at 2:33
Théophile
19.4k12946
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: Let $H$ (for half-life) be $1580$ years. Then if $f(t)$ is the remaining quantity in grams after a time $t$, we have:
$$begin{align}f(H) &= frac12\
f(2H) &= frac1{2^2}\
f(3H) &= frac1{2^3}, textrm{etc.}
end{align}$$
In general, we see that $f(xH) = 2^{-x}$. Now try to manipulate this to be able to work with the times you've been given.
answered Nov 23 at 2:33
Théophile
19.4k12946
Hint: Let $H$ (for half-life) be $1580$ years. Then if $f(t)$ is the remaining quantity in grams after a time $t$, we have:
$$begin{align}f(H) &= frac12\
f(2H) &= frac1{2^2}\
f(3H) &= frac1{2^3}, textrm{etc.}
end{align}$$
In general, we see that $f(xH) = 2^{-x}$. Now try to manipulate this to be able to work with the times you've been given.
answered Nov 23 at 2:33
Théophile
19.4k12946
answered Nov 23 at 2:33
Théophile
19.4k12946
answered Nov 23 at 2:33
Théophile
19.4k12946
answered Nov 23 at 2:33
Théophile
19.4k12946
19.4k12946
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up vote
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Let $D$ be the rate of decay. Then we have that:
$$e^{1580D}=frac 12$$
You can use this to find $D=frac{-ln(2)}{1580}$
Then you simply need to plug in $D$ for $e^{500D}$ etc
answered Nov 23 at 2:47
Rhys Hughes
4,6751327
add a comment |
up vote
0
down vote
Let $D$ be the rate of decay. Then we have that:
$$e^{1580D}=frac 12$$
You can use this to find $D=frac{-ln(2)}{1580}$
Then you simply need to plug in $D$ for $e^{500D}$ etc
answered Nov 23 at 2:47
Rhys Hughes
4,6751327
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $D$ be the rate of decay. Then we have that:
$$e^{1580D}=frac 12$$
You can use this to find $D=frac{-ln(2)}{1580}$
Then you simply need to plug in $D$ for $e^{500D}$ etc
answered Nov 23 at 2:47
Rhys Hughes
4,6751327
Let $D$ be the rate of decay. Then we have that:
$$e^{1580D}=frac 12$$
You can use this to find $D=frac{-ln(2)}{1580}$
Then you simply need to plug in $D$ for $e^{500D}$ etc
answered Nov 23 at 2:47
Rhys Hughes
4,6751327
answered Nov 23 at 2:47
Rhys Hughes
4,6751327
answered Nov 23 at 2:47
Rhys Hughes
4,6751327
answered Nov 23 at 2:47
Rhys Hughes
4,6751327
4,6751327
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