@JohnRennie Hello Sir :-)

@Jasmine hi :-)

@JohnRennie I had some doubt in nuclear physics

I'll help if I can. What's the question?

@JohnRennie what exactly occurs during alpha decay

It depends on what you mean by exactly occurs.

@JohnRennie Like how is the charge conserved

@Jasmine charge conservation just means the the charge of the decay product plus the charge of the alpha particle add up to the charge of the original nucleus.

Alpha particle has -2e charge, initially the atom wasnt charged so does that kean the new atom is doubly negative charged

i.e. the charge of the decay product is 2e less than the charge of the original nucleus, and the He nucleus carries away the missing 2e charge.

@Jasmine yes, though the decay is such a high energy that the electrons around the atom can be displaced. The decay produces a charged object, but how many electrons are left round that charged object will vary in a random way.

@JohnRennie Ok , but in books I never saw it being mentioned

We don't usually consider what happens to the electrons.

But if you start with a neutral block of an alpha emitter then in general that block will build up a negative charge as it decays.

@JohnRennie Ok

What is gamma decay then

@Jasmine If you have a hydrogen atom with the electron in a 2p orbital then the electron can decay down to the 1s orbital and emit a photon. Yes?

@JohnRennie yes

Well the protons and neutrons in a nucleus live in orbitals just like the electrons in an atom live in orbitals. Gamma decay happens when a nucleus starts in an excited state, like a hydrogen atom in a 2p state, and a proton or neutron in the nucleus decays to a lower energy orbital.

The photon emitted is the difference in energy between the two nuclear orbitals, and in nuclei the orbital energies are huge so the photon emitted is a gamma ray not visible light.

@JohnRennie so is it like in a gamma decay its the proton or neutron which participates

@Jasmine yes

The electrons surrounding the nucleus are not involved.

@JohnRennie Ok

Nuclear energies are so much higher than the energies of the electrons that in most cases you can just ignore the electrons. e.g. an isolated nucleus would decay in exactly the same way as a nucleus in a neutral atom.

If its given in the question that an atom undergoes beta - decay and subsequently emits gamma rays, how are we going to utilise the information that 'it undergoes gamma decay'

Beta decay is when a neutron in the nucleus decays to a proton and an electron. The electron zooms off leaving just the proton. But this often leaves the nucleus in an excited state i.e. the proton created by the beta decay is in a higher energy orbital than it could be.

So after the initial beta decay it's often the case that the proton then decays into a lower energy orbital and emits a gamma ray photon.

@JohnRennie Ok got it

One more doubt suppose X(a,b)->Y(c,d)+He(2,4)+ Energy

(a,b) denotes (no.of protons,neucleons)

In which form is the energy being released here

The energy goes into the kinetic energy of the decay products - mostly the kinetic energy of the alpha particle.

In particle reactions energy is released as either photons or the kinetic energy of the products. There isn't any other place for the energy to go.

@JohnRennie and how to calculate that energy, isnt it simply ∆mc^2 where ∆m is the mass difference of Y+He and X

@JohnRennie can it be in form of heat

@Jasmine yes, the energy is simply the mass lost times c^2.

In nuclear reactions mass is usually treated as the same as energy. In fact we usually measure particle masses in units of energy. For example any nuclear physicist will tell you that the mass of an electron is 511 keV or the mass of a proton/neutron is around a GeV.

So you just take the total energy before the decay and subtract off the rest mass energy of all the products. Then the difference is the energy released as a photon or KE.

@Jasmine heat is a collective property. A single particle doesn't have a temperature. You can only define a temperature when you have lots of particles in an assembly.

@JohnRennie ok

We need to find energy released during the process

Isnt it simply ∆mc^2 but probably we are required to find velocity of alpha particle and hence its Kinetic energy

And thats infact correct answer

Why can't we simply calculate ∆mc^2 here

Well you aren't told the masses of the X and Y nuclei, so you can't calculate the energy difference from the mass change. Note that the mass of the nucleus is not the same as the masses of the protons and neutrons inside it.

@JohnRennie Ok

@JohnRennie will that mean if the reaction happens many times then the heat energy can be released

For heat to have any meaning you normally require of order of a mole of the particles i.e. squillions of them!

@JohnRennie ok :-) understood

@JohnRennie Hello

@pi-π hi :-)

@JohnRennie Justify "Terminal voltage goes on increasing on decreasing the load current."

I need some context. What are we talking about? What physical system?

@JohnRennie Just an electric circuit.

What circuit?

Perhaps you could post a picture of the question.

@JohnRennie 1a

@pi-π Ah, I suspect that is referring to the voltage measured across the terminals of a battery, or other voltage source. This is the EMF minus the voltage drop across the internal resistance i.e. V = E - IR_i, where R_i is the internal resistance.

So as I increases V decreases and vice versa.

@JohnRennie Is there any physical explanation?

@pi-π Yes, all batteries have some internal resistance.

@JohnRennie Hello :-)

Small doubt

@Jasmine hi :-) What's the question?

@JohnRennie why cant we conserve mechanical energy just before and after explosion

In the explosion chemical energy is converted to mechanical energy. Energy is still conserved but only if you count all types of energy.

we conserve mechanical energy just before and after explosion:False

@JohnRennie we are talking about explosion during projectile motion

So initially the projectile is having some mechanical energy

Yes, but the explosion increases the total mechanical energy by converting chemical energy to mechanical energy.

Momentum is conserved, but mechanical energy is not.

@JohnRennie Oh ok got it !

I dont know the only thing that strikes me after reading energy is not conserved is work done by non conservative forces is non 0

I was trying to figure out how though

Uploading image here has become cumbersome! Either it doesn't upload or I need to compress it at times

This isn't really the same as a non-conservative force. That's normally a force like friction where energy is drained away as heat.

@JohnRennie How to apply work energy theorem here

In general you wouldn't solve questions like this using the work energy theorem because it's impossible to know how much of the chemical energy got converted to mechanical energy.

Typically you use conservation of momentum.

Have you got an example question we could look at?

@JohnRennie yes

@JohnRennie yes one second

OK, so which do you think are true?

Well its an analytical question

@JohnRennie A B

And D as well

Yes. I agree.

Yea thats the answer as well

D is less obvious, but if we assume the explosion happens very quickly the impulse delivered by the gravitational force will be small because $t$ is small.

@JohnRennie How is gravitational force impulsive here

Impulse is just force times time.

We generally call a force impulsive if we get a very large force acting for a very short time.

@JohnRennie yes.

Explosive forces must be large in magnitude

An explosion falls into that category because the actual explosion only lasts a very short time. The explosion delivers a large total impulse only because the forces are enormous!

By this I meant gravitational force is acting all along the motion ( for long time )

By impulsive force I mean large force that acts for small time

Suppose the explosion lasts a time $\Delta t$. Then the gravitational force delivers an impulse $\Delta p = mg\Delta t$ and this is not zero.

However it is much, much smaller than the impulse delivered in the same time by the explosive forces.

And that's what option D says.

@JohnRennie Oh! The question mentions during the explosion so ∆t is same for both

I got it @JohnRennie :-)

:-)

Good night :-)

Bye :-)

:-)