# Half life of Uranium question

## Homework Statement

I'm not sure if whether mass changes when atoms decay. Does emitting alpha/beta/gamma cause the atoms to lose mass? I don't know

## Homework Equations

Uranium-235 has a half life of 7.35x10^5 years and uranium 238 has a half life of 4.5x10^6 years. Compare the two isotopes of uranium, given that they both have the same number of atoms.

a) Which sample would have the greater weight?
b) which sample would have the greater number of decays per second?

## The Attempt at a Solution

For a) I'm not really sure but I think it would be the U-238 because it has the greater half life

For b) I'm not sure if this is right, if shorter half-life means greater decays per second. But I think it should be the U-235

## Answers and Replies

I like Serena
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Hi paperdoll!

For a) I'm not really sure but I think it would be the U-238 because it has the greater half life

Do you know what the number 238 represents?

For b) I'm not sure if this is right, if shorter half-life means greater decays per second. But I think it should be the U-235

What is the definition of half-life?

Hi paperdoll!

Do you know what the number 238 represents?

What is the definition of half-life?

Hi I like Serena, long time no see ^^ since I haven't posted here for a while :)

Yes, the number 238 represents the amount of protons and neutrons in the Uranium atom

okay, half life is the time it takes for an amount of a substance to decay down to half its original value.

I like Serena
Homework Helper
Hi I like Serena, long time no see ^^ since I haven't posted here for a while :)

Yep. I know!
Good that you remember me! :shy:

Yes, the number 238 represents the amount of protons and neutrons in the Uranium atom

Yes.
So which atom is heavier?

okay, half life is the time it takes for an amount of a substance to decay down to half its original value.

Right.
So suppose each sample has 2N atoms.
Then N atoms decay from 2N atoms in the corresponding half-time T.
How many decays is that per second?

Yep. I know!
Good that you remember me! :shy:

Yes.
So which atom is heavier?

Right.
So suppose each sample has 2N atoms.
Then N atoms decay from 2N atoms in the corresponding half-time T.
How many decays is that per second?

Yes, I have my final physics exams soon so I'm going through the past exam papers now :) that means more questions ahah
Oh! So the U-238 must be heavier since it has more neutrons
I'm not really sure about the decays per second. Does the decay per second stay constant throughout the entire decay process or does it slowly get smaller?

I like Serena
Homework Helper
Oh! So the U-238 must be heavier since it has more neutrons

Yep!

I'm not really sure about the decays per second. Does the decay per second stay constant throughout the entire decay process or does it slowly get smaller?

That's what I'm trying to get you to find out. :uhh:

If we start with 2N atoms, then after a half-time T, N atoms have decayed.
Can't you calculate the number of decays per second from that?

Yep!

That's what I'm trying to get you to find out. :uhh:

If we start with 2N atoms, then after a half-time T, N atoms have decayed.
Can't you calculate the number of decays per second from that?

would that be N/t decays per second then? o__O
that is, provided that the "t" value is measured in seconds I think :uhh:

I like Serena
Homework Helper
would that be N/t decays per second then? o__O
that is, provided that the "t" value is measured in seconds I think :uhh:

Yep. :)
Note that if t is not in seconds, you can simply convert it to seconds.

Now to answer your last question first - does it decrease in time or not?

Yep. :)
Note that if t is not in seconds, you can simply convert it to seconds.

Now to answer your last question first - does it decrease in time or not?

um...so the number of decays per second should decrease after time right? because nuclear decay is an exponential curve, so eventually there will be zero decays per second I think

I like Serena
Homework Helper
um...so the number of decays per second should decrease after time right? because nuclear decay is an exponential curve, so eventually there will be zero decays per second I think

Right! ;)

Let's put it this way.

If you start with 2N atoms, you've got N/T decays per second.

After a time T you'd be left with N atoms.
In the next period T, N/2 atoms would decay.
So you have (N/2)/T decays per second.
That's less, isn't it?

It is indeed a downward exponential curve.

Okay.
So if we look at the two samples, of each 2N atoms with half-times T1 and T2.
What are their respective decays per second?
And which is the greater?

Right! ;)

Let's put it this way.

If you start with 2N atoms, you've got N/T decays per second.

After a time T you'd be left with N atoms.
In the next period T, N/2 atoms would decay.
So you have (N/2)/T decays per second.
That's less, isn't it?

It is indeed a downward exponential curve.

Okay.
So if we look at the two samples, of each 2N atoms with half-times T1 and T2.
What are their respective decays per second?
And which is the greater?

okay hmm, so if I let T1 be U-235= that means 7.35x10^5 years
and let T2 be U-238 = that means 4.5x10^6 years

so the rate of decay will be N/T(1 or 2)

so it is N/(7.35x10^5) and N/(4.5x10^6 years)

so the one with the greater number of decays per second will be N/(7.35x10^5) which is T1 which is U-235

I think I get it now Thank you I like Serena ^^ you made things extremely clear now

I like Serena
Homework Helper
okay hmm, so if I let T1 be U-235= that means 7.35x10^5 years
and let T2 be U-238 = that means 4.5x10^6 years

so the rate of decay will be N/T(1 or 2)

so it is N/(7.35x10^5) and N/(4.5x10^6 years)

so the one with the greater number of decays per second will be N/(7.35x10^5) which is T1 which is U-235

I think I get it now Thank you I like Serena ^^ you made things extremely clear now

You're welcome!

"I'm not sure if this is right, if shorter half-life means greater decays per second."

That was true you know, but you were not sure...
I hope you don't mind that I took you the long route.

You're welcome!