@JohnRennie: Hi sir. Good morning :-)

@GuruVishnu hi :-)

If you find time, could you clarify this doubt, sir? :

In a nuclear reaction can we say that the energy released is equal to the difference in the binding energies of the reactants and products? I understand that the energy released is due to the mass defect as reactants turn to products.

It's (2m(alpha)-m(lithium)) for subtracting the initial and final binding energies and

it's (m(lithium)+m(proton)-2m(alpha)) when we find the mass defect and find the energy due to the conversion.

Clearly the first method doesn't work.

Could you tell why we have inconsistency between the two methods?

Hmm ...

Did you get how I obtained that expression, sir?

I think this should work ...

You're saying the binding energy on the left is $E_{Li} = 3m_p + 4m_n - m_{Li}$ and on the right it's $2E_\alpha = 4m_p + 4m_n - 2m_\alpha$. Then the $m_p$ and $m_n$ cancel.

@JohnRennie On the left I think we must have another $m_p$ term as a proton is another reactant.

So it must be something like: $E_{Li}=4m_p+4m_n-m_{Li}$

Take your second approach of subtracting the masses and write $m_{Li} = 3m_p + 4m_n - E_{Li}$ and $m_\alpha = 2m_p + 2m_n - E_\alpha$.

You're going to get the same equation.

Ok sir. Let me try. In this:

The LHS is the initial energy and not only specific to lithium. I made a small mistake while typing it.

It must be $E_{initial}$ instead of $E_{Li}$

Your $E_{initial}$ is the energy you'd need to separate a lithium nucleus into the individual nucleons. You could do the same on the right hand side to find the energy released.

However, I think this doesn't match with the difference of the masses of the reactants and products - $m_{Li}+m_{p}-2m_{He}$.

If I use numerical values,

the first one turns out to be 920.9452 MeV

and the second one is 16.828756 MeV.

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I need to work now for about half an hour. I'm not sure I understand what you are doing, but your result is obviously out by the energy of one proton.

Ok sir.

@GuruVishnu hi, I've finished work now so I'm free for the rest of the morning.

Can we start again because I've lost track of what you are trying to do.

@JohnRennie Hi.

@JohnRennie Ok sir.

@GuruVishnu hi

Let's forget everything about that question. I'm going to explain from the beginning, how I got that one and so on.

Ok?

OK ...

Please wait till I type EOM - End of message.

I got this doubt when I tried to solve the following question in a different way:

The method which gives the correct answer is: finding the difference between the mass of the reactants and the products in unified mass (u) and then multiply with $931$ to obtain the energy in MeV (Mega electron volts).

Mathematically,

$E=[m_{Li}+m_p-2m_{\alpha}]c^2$

Now, let's discuss about the alternate method I used which in fact gave the incorrect result.

We know that both the reactants and the products are made of protons and neutrons. When we form the reactants from isolated nucleons, some energy is released which is nothing but the binding energies of the reactants. Same is the case of products.

My assertion is: the energy released in the reaction must be equal to the difference between the initial and the final energies, i.e., between the reactants and the products.

So we get,

Comparing the result through the first method - $E=[m_{Li}+m_p-2m_{\alpha}]c^2$ and the second method - $U_i-U_f=[2m_{\alpha}-m_{Li}]c^2$, clearly there is a discrepancy. Also this inconsistency is verified by substituting the numerical values which vary so drastically.

The quantitative results are:

First method: 16.828756 MeV

Second method: 920.9452 MeV

I want to know why the two results don't match? Do I need to know something else? Did I go wrong anywhere (as far as I have thought, I'm unable to figure it out)?

EOM

Let me try and restate what I think your second approach is doing:

We start with four protons and four neutrons and we end with four protons and four neutrons. So the only thing that can have changed is the binding energy.

Yes sir.

That means if the initial binding energy is $E_i$ and the final binding energy is $E_f$ the energy released in the reaction must be $\Delta E = E_f - E_i$

OK so far?

Yes sir. In the above block of text, I have used $U$ instead of $E$ (just a different notation).

The final binding energy is easy because we have only alpha particles so the total binding energy is $2E_\alpha$ where $E_\alpha = 2m_p + 2m_n - m_\alpha$.

OK so far?

Ok sir.

On the left side we have two different objects, a lithium nucleus and a proton. The total binding energy is the sum of the binding energy of a lithium nucleus plus the sum of the binding energy of a proton. Yes?

@JohnRennie Yes sir. The binding energy of the lithium is straight forward. Whereas it's zero for a proton. However, I included the $+m_p$ term to account for it's rest mass energy.

Yes, and there is your mistake. We are trying to calculate the binding energy not the rest mass energy. We already took care of the rest mass energies when we balanced out the four protons and four neutrons on each side.

Ah. Ok sir. Now I can see where I went wrong.

Cool :-)

Thank you sir :-)

I'm off to answer a question on quantum mechanics in another room, but I'll be around for the rest of the morning if you want to ask more.

Ok sir.

Sir, in some answer on the main site I noticed a neutrino on the product side when transferred to the reactant side it becomes an antineutrino and vice versa. So is it something like $\nu=-{\nu}^{-}$? And is the case same for a positron and an electron?

I don't remember where I saw that kind of transformation.

Yes

In effect you add an antineutrino to both sides and the $\nu$ and $\bar\nu$ cancel.

Ok sir. So both an electron capture reaction and beta plus reaction are merely the same?

Or do they take place simultaneously?

Some care is needed ...

Let's start with the electron capture:

$$ p + e \to n + \nu_e $$

Yes sir.

Now we can add a positron to both sides and allow the electron and positron to cancel on the left side. Then we get:

$$ p + 2\gamma \to n + e^+ + \nu_e $$

Because the electron and positron annihilate to two photons so they can be treated as equaivalent to two photons.

Where did $\gamma$ come from?

@JohnRennie I see. I have to learn about quantum "annihilation" before I question this further. I remember, some time ago, we came to the same concept.

@GuruVishnu the electron and positron both have the same energy of $m_e c^2$ so they can't just vanish or energy wouldn't be conserved.

@JohnRennie Ok sir. So $e^-+e^+=\text{energy}$ instead of $0$. Did I get it?

Yes

Although writing energy is a poor way to put it. The electron and positron convert into two new particles i.e. two photons.

Photons are not energy - they are massless particles.

@JohnRennie Two photons of equal energy?

Yes.

The reason two photons are produced is because momentum cannot be conserved otherwise.

Consider the reaction in the centre of momentum frame i.e. total momentum is zero.

@JohnRennie Ok sir. Now I can see why they must be of the same energy.

So initially the electron and positron have equal and opposite momenta.

The photon momentum is $h/\lambda$

So the only way for momentum to be conserved is if we get two photons with equal and opposite momenta $+h/\lambda$ and $-h/\lambda$.

Ok sir. If the initial momentum is non-zero, say we hit an electron on a fixed positron, again, will two photons be emitted? If yes, I think the two photons will have different energies and hence different momenta.

There is always a centre of momentum frame, so there must always be two photons produced.

In a frame that is not the COM frame the photons will not have equal and opposite momenta.

The photon emitted in the direction of light will be blue shifted and the other be red shifted. Is this conclusion correct?

In general the motion relative to the COM frame will change the direction of the photon momenta, and it will blue shift one of the photons and red shift the other.

@GuruVishnu yes

@JohnRennie Ok sir. I'm slightly confused, why not the opposite take place - red shifted in the prograde direction and blue shifted in the retrograde direction?

Let's consider an example. I'll draw a diagram:

Suppose we have a stationary positron (red) being hit my a moving electron (blue).

Ok sir.

So if we take the positive x direction to be right we the total momentum is $+P_x$

Yes sir.

On collision the particles annihilate. We'll assume the photons get produced along the line of motion - they can be produced in any direction but we'll assume along the line of motion for simplicity.

Ok sir.

Then the momentum of the right moving photon is $+h/\lambda_r$ and the momentum of the left moving photon is $-h/\lambda_l$.

So we require $+h/\lambda_r - h/\lambda_l = P_x$

@JohnRennie Yes sir. This shows why is $\lambda_l>\lambda_r$

That immediately tells you $\lambda_r < \lambda_l$ so the right moving photon is blue shifted amd the left moving photon is red shifted.

Fine sir. I understood this.

Though it could look like this:

i.e. the photons emitted at equal angles to the direction of travel. In that case their wavelengths would be equal.

@GuruVishnu how did you get confused?

@JohnRennie Ok sir. I understood everything we discussed so far. But is it necessary to have two photons after collision? Is this a property of "annihilation" (quantum mechanics)?

You can't have one photon because with one photon there is no way to conserve momentum. So the question is could you have three, four, five, etc photons. Yes?

@JohnRennie Yes sir.

> the photons emitted at equal angles

I understand why the photons must be emitted at equal angles, else the momentum along the vertical will not be conserved.

But, in the final case, why can't the equal angles be equal to zero. Are there any other values, these angles can't take?

And the answer is that yes you could have three or more photons produced from an $e \bar{e}$ annihilation, but when you calculate the probability of this happening it turns out to be much, much smaller than the probability for two photons.

With massive particles the momentum is proportional to $v$ and the energy to $v^2$, so in general any angles are allowed because you can always find a combination of angle and velocity that satisifes both conditions.

But for photons $E = pc = hc/\lambda$ so the energy and momentum are both inversely proportional to the wavelength.

This restricts the possibilities for the angle of emission.

@JohnRennie Ok sir. But I don't see any violation of conservation of momentum if only one photon is emitted along the direction of initial momentum. Anyway, if two photons must be emitted, I've thought of another reason based on change of reference frame. If we imagine ourselves to be in an inertial frame where the momentum of the two particles is initially zero, two photons must be emitted and they must have equal energies.

Emission of two photons is invariant of the reference frame. So even in the original reference frame two photons must be emitted, but they will be blue and red shifted accordingly.

Could you tell whether my reasoning is correct?

> Ok sir. But I don't see any violation of conservation of momentum if only one photon is emitted along the direction of initial momentum.

So what would this look like in the centre of momentum frame?

Yes sir. I realise my mistake. So our observation (emission of two photons) must not depend on the reference frame we choose - something like, one photon in one frame and two in the other frame is not permissible. Am I right?

Correct. Changing the frame is just a choice of coordinates i.e. just mathematics. It cannot change things like the number of photons created.

Ok sir. Could you comment on my reasoning in my previous message? Now I feel it's also correct.

"...if two photons must be emitted, I've thought of another reason... ...and red shifted accordingly."

Is this correct?

Yes

Just curious, in which reference frame will the total energy $E$ be distributed as $0$ and $E$ for the retrograde and prograde photons? Is that the frame which travels in the speed of light in the direction of the initial momentum?

Or is that the frame travelling with the speed of light in the direction opposite to the initial momentum?

@GuruVishnu there is no frame in which the energy of the red shifted photon is exactly zero because that would require travelling at the speed of light, and in that frame our electron and positron would have to be travelling at the speed of light. And massive particles cannot travel at the speed of light.

But it possible to travel arbitrarily close to speed of light, and in that frame the energy of the red shifted photon can come arbitrarily close to zero.

@JohnRennie Ah. Nice. And this also explains why there must necessarily be two photons. A proof by contradiction.

It's good fun this physics thing :-)

Thank you for your help and time sir :-) Colliding quantum particles is more fun than I thought initially. People at particle accelerators probably enjoy their lives.

I would guess the people working at CERN really enjoy it. It has to be a dream job :-)

:-)

@JohnRennie: Hi sir. Are you free now?

@GuruVishnu hi, yes I'm for a bit.

Ok sir.

Without prior knowledge on the reactions, is it possible to answer the following question? :

> Lithium (Z=3) has two stable isotopes of mass numbers 6 and 7. When neutrons are bombarded on lithium sample, electrons and alpha particles are ejected. Write down the nuclear processes taking place.

I just converted Li-6 and 7 to electrons and hydrogen isotopes, but the final answer is more different than this.

Is there any criterion for neutron addition in such reactions?

I think you can suggest plausible options. For example electrons can only be produced by beta decay, so the most likely reaction is:

$$ {}^7Li + n \to {}^8Li \to {}^8Be + e + \nu $$

Yes sir. That was how the author approached the first step.

I guess this could happen with Li-6 as well. That would produce Be-7.

As for alpha particles, the only plausible reaction would be something like:

Yes sir. Later he broke beryllium into two alpha particles. But may I know how did you know to add a neutron to the nucleus? Any logic? Or is that based on experimental results?

$$ {}^7Li + n \to {}^8Li \to 2 \alpha $$

Well it says you are bombarding a lithium nucleus with neutrons, so the initial reaction is going to be that the nucleus absorbs the neutron so its atomic number remains the same and its mass increases by one.

@JohnRennie Ok sir. He did it something like this: $$^8 Be\to 2 \alpha$$ I don't know how to decide when a neutron will add to a nucleus and when will it cause a decay.

It's impossible to predict what will happen but you can say what might happen.

It's very unlikely that more than one neutron would be absorbed by the nucleus, so the intermediate step will be either Li-7 or Li-8 depending on whether you started with Li-6 or Li-7.

Li-7 is stable so it doesn't seem likely anything much will happen to it.

Ok sir. I came up with the following reaction: $$^7 Li+^1 n\to e^-+^4 He+^3 H$$ could you tell whether this one is possible or not. If not possible, could you give any reasons for it?

That would require Li-8 to decay to an alpha, a tritium nucleus and an electron.

Wait, the charge doesn't balance in your reaction.

On the left the total charge is +3 from the three protons and on the right it's +2 from three protons and one electron.

I'm only counting the nuclear charges and ignoring the valence electrons.

Ok sir.

@JohnRennie I agree with your point sir.

$$ ^7 Li+^1 n\to e^-+^4 He+^3 He $$

would be possible but unlikely as alpha particles don't beta decay.

Ok sir. But why do we need to consider "alpha particles don't beta decay" here?

Oh wait, that's not right.

Ignore me.

Ok sir. No worries.

I'm going to head off in search of lunch I think. I'm getting tired now.

Ok sir. Good bye. Take care :-)

Bye :-)