Radium has a half-life of 1580 years. If a sample contains 1g of radium now, how much radium will there be:
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
add a comment |
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
add a comment |
up vote
0
down vote
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
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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
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
4,6731327
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009901%2fradium-has-a-half-life-of-1580-years-if-a-sample-contains-1g-of-radium-now-how%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
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
add a comment |
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
4,6731327
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
4,6731327
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
4,6731327
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
4,6731327
answered Nov 23 at 2:47
Rhys Hughes
4,6731327
answered Nov 23 at 2:47
Rhys Hughes
4,6731327
answered Nov 23 at 2:47
Rhys Hughes
4,6731327
4,6731327
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009901%2fradium-has-a-half-life-of-1580-years-if-a-sample-contains-1g-of-radium-now-how%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
