Kinetic energy of alpha particle

[Raghav Gupta]

Homework Statement

The kinetic energy (in keV) of the alpha particle, when the nucleus ₈₄²¹⁰Po at rest undergoes alpha decay, is
A 5319
B 5422
C 5707
D 5818

Homework Equations

$E = mc^2$

Kinetic energy $= mv^2/2$

The Attempt at a Solution

alpha particle is emission of helium nucleus
₈₄²¹⁰Po → ₂⁴He + ₈₂²⁰⁶Pb
Now E = mc², what mass we will take?

 

[Simon Bridge]

What masses have you got? List them.

 

[Raghav Gupta]

What masses have you got? List them.

Mass of polonium is 210u , of helium is 4u and of Pb is 206u, is that correct, seeing the superscript of elements?

 

[Simon Bridge]

Did you check those masses by looking them up?
So how much energy (using E=mc^2) did you start with, and how much of that energy ended up as mass after the alpha decay?

(Use $E_u=uc^2$)

 

[Raghav Gupta]

The difference is zero.
Earlier mass was 210u and afterwards also 206u + 4u= 210u. So no energy released?

 

[Simon Bridge]

If all the initial energy goes to the daughter masses, with nothing left over for kinetic energy, then what is the kinetic energy equal to?

But here is the thing: is the polonium mass the same as the mass of 84 protons and 126 neutrons (and 84 electrons ... but we are thinking nuclear)?

 

[Raghav Gupta]

If all the initial energy goes to the daughter masses, with nothing left over for kinetic energy, then what is the kinetic energy equal to?

But here is the thing: is the polonium mass the same as the mass of 84 protons and 126 neutrons (and 84 electrons ... but we are thinking nuclear)?

The K.E is zero. But it is not given in options.

Yeah, it should be.

 

[BvU]

Hello again, Raghav and Simon,

What Simon means is you have to look up these masses to high precision: 5 or 6 MeV isn't much compared to these >200 u, so you'll have to go into detail.

Then again: in your tests, do you have a table with masses available? or do you have to learn them all by heart -- in 8 or 9 figures, to judge from this exercise...

 

[Raghav Gupta]

From where that 5 or 6 MeV came?
Isn't ΔE = 931.5 MeV for 1 u?

Mass of Pb here = 205.974455u
Mass of He here = 4.002603
Mass of Po here = 209.982876

 

[Raghav Gupta]

Then again: in your tests, do you have a table with masses available? or do you have to learn them all by heart -- in 8 or 9 figures, to judge from this exercise...

Oh, a table was given at the back of the page.
I did't know that we have to look at it.
If it would be asked in test, then I think they will definitely give us a table.

 

[BvU]

I got the 5 or 6 MeV from your post #1:

A 5319 keV
B 5422
C 5707
D 5818

OK, now that you have better masses (what about the $\alpha$ ?), do the math again.

 

[Raghav Gupta]

I got the 5 or 6 MeV from your post #1:

A 5319 keV
B 5422
C 5707
D 5818

OK, now that you have better masses (what about the $\alpha$ ?), do the math again.

I am not getting it,
How you got 5 MeV from post 1
1u mass has energy equivalent to 931.5 MeV, then α particle will have 931.5 x 4 MeV energy , how 5 MeV ?

 

[Raghav Gupta]

But here is the thing: is the polonium mass the same as the mass of 84 protons and 126 neutrons (and 84 electrons ... but we are thinking nuclear)?

So I see, polonium has less mass then individual atoms and neutrons as some mass goes for binding energy.

 

[BvU]

I got 5-6 MeV because that was the range of the possible answers. We know one of them must be correct, right ?

So I see, polonium has less mass than the sum of the masses of the individual atoms protons and neutrons as some mass goes for binding energy.

Yes, and lead also less, to a greater extent. And He has more.

M(in u) $-$ A is called mass excess (negative for ²¹⁰Po and ²⁰⁶Pb, positive for ⁴He). Your little table on the back doesn't list excess masses but actual masses, which is fine too.

Now do the (very simple) math following from energy conservation and mass/energy equivalence. Since we work in u (or Daltons), we need your "1u mass has energy equivalent to 931.5 MeV", not c² as conversion from u to E_kin. By the time we have that, there is another nasty twist, which we will discuss once you have a value of the right order of magnitude.

 

[Raghav Gupta]

How, He has + ve mass excess? How it can have more mass then the sum of individual masses of neutrons and protons it has? Doesn't He has binding energy?

I subtracted the mass of polonium given in table from sum of lead and helium , and that excess mass I used in energy - mass equivalence theorem and got energy 5403 KeV approximately by hand and from calculator 5420 KeV( which we are not allowed in exams), almost the option B.

Is it the kinetic energy or if not then what energy?( You are saying that here is a nasty twist. )

 

[BvU]

Good. Done with energy conservation. You got past the 5818 trap (wrong units) and are now at about 5420 (litterature 5407 keV. It goes by the name Q). Time to check yet another conservation law - one that's always valid. At this point what you have is the total kinetic energy, right ? Remember the problem statement said something speific about the ²¹⁰ Po ?

 

[BvU]

How, He has + ve mass excess? How it can have more mass then the sum of individual masses of neutrons and protons it has? Doesn't He has binding energy?

from Wang et al :

  N  Z  Elt.   Mass excess                 Binding energy            Atomic mass
                 (keV)                       per nucleon (keV)           μu

126 84 Po      -15952.7         1.2          7834.344 0.006      209982874.1         1.3

124 82 Pb      -23785.1         1.2          7875.359 0.006      205974465.7         1.3

  2  2 H         2424.91561     0.00006      7073.915 0.000        4002603.25413   0.00006

And from the excess mass in keV you see the 5407 keV coming out straightforward.

 

[Simon Bridge]

I got distracted... lessee... where are we... oh you've discovered that the masses are not so easily calculated, well done :)

How, He has + ve mass excess? How it can have more mass then the sum of individual masses of neutrons and protons it has? Doesn't He has binding energy?

The mass excess is +ve because He has binding energy.

Helium actual mass is 4.002602u (look it up), but it has atomic number 4 so a mass excess (by definition) of

(4.002602 - 4)u = +0.002602u​

... a positive number.

That's 2 protons and two neutrons ... the combined mass of those nucleons when free would be:
2x1.007276u + 2x1.008645u = 4.031843u, which, as you observe, is bigger than the actual mass of the nucleons when bound up as an alpha particle.

It's usually easier to handle mass excesses than actual masses.

 

[Raghav Gupta]

Good. Done with energy conservation. You got past the 5818 trap (wrong units) and are now at about 5420 (litterature 5407 keV. It goes by the name Q). Time to check yet another conservation law - one that's always valid. At this point what you have is the total kinetic energy, right ? Remember the problem statement said something speific about the ²¹⁰ Po ?

Not understanding the bold part. What is Q? And how 5420 is 5407 in literature?

Yeah, we can do the conservation of momentum. And as Po is at rest.
m_alphav_alpha = m_Leadv_lead
Also,
$\sqrt{2m_{alpha}KE_1} = \sqrt{2m_{lead}KE_2}$

from Wang et al :

  N  Z  Elt.   Mass excess                 Binding energy            Atomic mass
                 (keV)                       per nucleon (keV)           μu

126 84 Po      -15952.7         1.2          7834.344 0.006      209982874.1         1.3

124 82 Pb      -23785.1         1.2          7875.359 0.006      205974465.7         1.3

  2  2 H         2424.91561     0.00006      7073.915 0.000        4002603.25413   0.00006

And from the excess mass in keV you see the 5407 keV coming out straightforward.

How 5407 is coming straight forward?

I got distracted... lessee... where are we... oh you've discovered that the masses are not so easily calculated, well done :)
The mass excess is +ve because He has binding energy.

Helium actual mass is 4.002602u (look it up), but it has atomic number 4 so a mass excess (by definition) of

(4.002602 - 4)u = +0.002602u​

... a positive number.

That's 2 protons and two neutrons ... the combined mass of those nucleons when free would be:
2x1.007276u + 2x1.008645u = 4.031843u, which, as you observe, is bigger than the actual mass of the nucleons when bound up as an alpha particle.

It's usually easier to handle mass excesses than actual masses.

Thanks, that explains my that question pretty well.

 

[BvU]

How 5407 is coming straight forward?

Now you disappoint me. I thought you could work out mass excess before = mass excess after all by yourself especially if the numbers are in front of you. A little juggling would tell you how. But let me show you:

       Before  after
 
       -15952.70  -23785.10
                    2424.92
                    5407.48
total  -15952.70  -15952.70

m_alphav_alpha = m_Leadv_lead is good enough to find out how to distribute the kinetic energy over $\alpha$ and Pb.

Your turn

--

 

[BvU]

Oh, and Q is just a name for the decay energy . (here too)

 

[Raghav Gupta]

Now you disappoint me. I thought you could work out mass excess before = mass excess after all by yourself especially if the numbers are in front of you. A little juggling would tell you how. But let me show you:

       Before  after

       -15952.70  -23785.10
                    2424.92
                    5407.48
total  -15952.70  -15952.70

m_alphav_alpha = m_Leadv_lead is good enough to find out how to distribute the kinetic energy over $\alpha$ and Pb.

Your turn

--

Why mass excess before = mass excess after?
Numbers are showing that but what is the reason for that. ( Sorry, I'm doing these type of problems first time.)
And that energy which I have calculated 5420 KeV, is it wrong?
Is the correct energy 5407 KeV?

We also don't know the velocities of lead and alpha particle in momentum conservation.
And if we use that mass figures upto 6 digits after decimal, wouldn't it be difficult dividing them without calculator?

 

[BvU]

Why mass excess before = mass excess after

That is simply conservation of energy (in the E =mc² form) where the 210 u are subtracted left and right.

Your 5420 keV is just fine (and closer to the tempting 5422 in answer B. Remember that you were given the masses).

We also don't know the velocities of lead and alpha particle in momentum conservation.

The exercise asks for a kinetic energy of the alpha. All we need is the ratio of the velocities, and that follows from momentum conservation. That ratio can be used in an expression for the ratio of the kinetic energies. You know the sum (this Q = 5420 keV), so then you have the $T_\alpha$. Try it!

figures upto 6 digits after decimal, wouldn't it be difficult dividing them without calculator

Not really.

I wanted to postpone this story until you figured out the correct answer, but never mind:
What you need to high accuracy is the difference (this value of Q). If you are given 205.974455, 209.982876 and 4.002603, you want to make sure you get all (4) digits from the difference, 5818 μu. (Which happens to be answer D -- no coincidence). No multiplication, just a simple subtraction, to be done on paper with ease.

Next step is to go from μu to keV/c², the 0.9314941 . So roughly 7% less. This means you can discard answer D and also C, but you still need some accuracy to distinguish between A and B. A and B differ 2%, so 1% should be enough already. The 0.93 = 100% - 7% brings you to answer B (again, no coincidence).

In the last step you need to make a division, namely by approximately 1.02 (You still have to find out why). That division by 1.02 is the same as multiplying with 0.98, so subtracting 2%. Bingo ! (But you knew the answer already ?)

The 5407 is from here , they follow Wang et al.. Accuracy is pretty good: 1.2 keV in Po and Pb​

 

[Raghav Gupta]

Okay, I just want to ask first that what will be the masses of lead and alpha particle when we apply momentum conservation and find ratio of velocities.
Can we take 206 u and 4 u respectively although they are not their actual masses but atomic mass?

 

[BvU]

Good enough. Don't ask for every step: try something and if it doesn't work try something else.

 

[Raghav Gupta]

Not understanding the bold part. What is Q? And how 5420 is 5407 in literature?

Yeah, we can do the conservation of momentum. And as Po is at rest.
m_alphav_alpha = m_Leadv_lead
Also,
$\sqrt{2m_{alpha}KE_1} = \sqrt{2m_{lead}KE_2}$

How 5407 is coming straight forward?

I think you missed the red part in my some earlier post?
So KE₁ + KE₂ = 5420 KeV ---1)
$\sqrt{2m_{alpha}KE_1} = \sqrt{2m_{lead}KE_2}$

$⇒ KE_2 = 2KE_1/103$
Finally substituting in 1),
We get,
$105KE_1/103 = 5420$
So, definitely answer will be less then 5420 which is option A
Doing it with hand fast 105/103 and then dividing 5420 will be difficult as you were stating, but we know that the answer will be less then 5420.

[Edit - add ] Why we take here lead mass as 206 u and alpha particle as 4u?
If lead has sum of protons and neutrons 206, then it's mass should not be 206 u but 206.something u
as proton mass is not 1u but 1.007.. u
Also there is actual mass which we take by mass spectrometer.
3 types of masses!

 

[BvU]

Didn't miss it, just found it too complicated; matter of taste.

$m_1 v_1 = m_2 v_2 \Rightarrow m_1 v_1^2 / m_2 v_2^2 = m_2/m_1,$

${\tfrac 1 2} (m_1 v_1^2 + m_2 v_2^2) = 5407 \Rightarrow {\tfrac 1 2}m_1 v_1^2 / 5407 = 1/ (1+m_2/m_1)$

RE accuracy required read post #23.

 

[Raghav Gupta]

Didn't miss it, just found it too complicated; matter of taste.

$m_1 v_1 = m_2 v_2 \Rightarrow m_1 v_1^2 / m_2 v_2^2 = m_2/m_1,$

${\tfrac 1 2} (m_1 v_1^2 + m_2 v_2^2) = 5407 \Rightarrow {\tfrac 1 2}m_1 v_1^2 / 5407 = 1/ (1+m_2/m_1)$

RE accuracy required read post #23.

Okay, I am not understanding in post 23 that why accuracy is required?
When we are dividing 5420 KeV by 1.02 surely we will get answer less then 5422. and only option that I see here is option A which is 5319 KeV . Reason of dividing by 1.02 I found out from equations in post 26.

[Edit -add]
In case if you have not seen my edit add of post 26
Why we take here lead mass as 206 u and alpha particle as 4u?
If lead has sum of protons and neutrons 206, then it's mass should not be 206 u but 206.something u
as proton mass is not 1u but 1.007.. u
Also there is actual mass which we take by mass spectrometer.
3 types of masses!

 

[BvU]

You are right, it's the same equation -- hence the same result.
And indeed, I am trying to say that for the step from 5407 to 5319 there is no need for high accuracy. So taking 206 / (4 + 206) is good enough.

There really aren't that many masses. The nucleon number isn't a mass. There is basically only the rest mass m₀ that can be expressed in kg, MeV/c² or u ( 205.9744657 for ²⁰⁶Pb ). When the atom is moving, you have the relativistic mass $E = \gamma m_0 c^2$ with $\displaystyle \gamma = {1\over \sqrt { 1-{v^2\over c^2}}}$ only depending on $|v|$, so not tabulated or anything.

 

[Raghav Gupta]

Thanks BvU for the so much help.
Also thanks to Simon.