# Kinetic energy of alpha particle

## 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?

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Simon Bridge
Homework Helper
What masses have you got? List them.

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
Homework Helper
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 $E_u=uc^2$)

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

Simon Bridge
Homework Helper
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)?

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
Homework Helper
2019 Award
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...

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

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
Homework Helper
2019 Award
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 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 ?

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
Homework Helper
2019 Award
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 Ekin. 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. )

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BvU
Homework Helper
2019 Award
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 ?

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BvU
Homework Helper
2019 Award
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 :

Code:
  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
Homework Helper
I got distracted... lessee... where are we... oh you've discovered that the masses are not so easily calculated, well done :)
Raghav said:
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.
Also,
$\sqrt{2m_{alpha}KE_1} = \sqrt{2m_{lead}KE_2}$
from Wang et al :

Code:
  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
Homework Helper
2019 Award
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:
Code:
       Before  after

-15952.70  -23785.10
2424.92
5407.48
total  -15952.70  -15952.70
malphavalpha = mLeadvlead is good enough to find out how to distribute the kinetic energy over $\alpha$ and Pb.

--

BvU
Homework Helper
2019 Award
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:
Code:
       Before  after

-15952.70  -23785.10
2424.92
5407.48
total  -15952.70  -15952.70
malphavalpha = mLeadvlead is good enough to find out how to distribute the kinetic energy over $\alpha$ and Pb.

--
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
Homework Helper
2019 Award
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​

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