Expected energy of a beta particle

[Richie Smash]

Homework Statement

Nitrogen-17 emits a Beta particle to form an isotope of Oxygen.
Use the following data to find the expected energy of the Beta particle.
Mass of N-17 = 17.00845 u
Mass of O-17 = 16.99913 u
Mass of Beta particle = 0.00055 u
Unified mass unit (u) = 1.66 x 10^-27kg
Speed of light = 3.00 x 10⁸m s^-1

Homework Equations

E=mc²

The Attempt at a Solution

I am aware that the energy can be calculated by E = mc²
But they have 3 different masses given, I'm unaware if I have to use simply the mass given for the Beta particle or have to do some calculation first...

 

[Gigaz]

Whoever wrote this exercise is either clueless or wants to teach something about Beta decay later. Essentially the idea here is just to remember conservation of mass/energy. Everything that is missing goes into kinetic energy.
Here's the problem with the exercise: Actually Beta decay always involves a neutrino. So what you get is just the maximum kinetic energy of the Beta.

 

[Richie Smash]

AH yes, I did read that in a textbook, however I must answer this question as it's from a practice exam, and I'm unsure as to why they gave me all those masses, even it if it is to calculate the kinetic energy as you say, are all those masses just to confuse me?

 

[Gigaz]

AH yes, I did read that in a textbook, however I must answer this question as it's from a practice exam, and I'm unsure as to why they gave me all those masses, even it if it is to calculate the kinetic energy as you say, are all those masses just to confuse me?

You see, the interpretation of E=mc^2 is extremely simple. Mass and energy are the same thing, just with different units, hence the factor. You also need to understand conservation of energy/conservation of mass (both terms equivalent) That means that the mass/energy before the decay must be the same as the mass/energy after the decay.

 

[PeroK]

Homework Statement

Nitrogen-17 emits a Beta particle to form an isotope of Oxygen.
Use the following data to find the expected energy of the Beta particle.
Mass of N-17 = 17.00845 u
Mass of O-17 = 16.99913 u
Mass of Beta particle = 0.00055 u
Unified mass unit (u) = 1.66 x 10^-27kg
Speed of light = 3.00 x 10⁸m s^-1

Homework Equations

E=mc²

The Attempt at a Solution

I am aware that the energy can be calculated by E = mc²
But they have 3 different masses given, I'm unaware if I have to use simply the mass given for the Beta particle or have to do some calculation first...

If you start with N-17 and you end up with O-17 and a beta particle, then what is missing?

 

[Richie Smash]

If you start with N-17 and you end up with O-17 and a beta particle, then what is missing?

The mass deficit?

 

[PeroK]

The mass deficit?

Exactly.

 

[Richie Smash]

Ok well in that case,
I will do this:
16.99913+0.00055 = 16.99968 u

SO mass deficit = 17.00845-16.99968 = 0.00877 u

Mass deficit in kg = 1.66 x 10^-27*0.00877 =1.45582 x 10^-29kg

So now Energy released would be
(3x10⁸)² * 1.45582 x 10^-29kg
E = 1.310238 x 10^-10J

However something is borthing me, is this the expected energy of the beta particle as well, or is that another separate calculation?

 

[PeroK]

Ok well in that case,
I will do this:
16.99913+0.00055 = 16.99968 u

SO mass deficit = 17.00845-16.99968 = 0.00877 u

Mass deficit in kg = 1.66 x 10^-27*0.00877 =1.45582 x 10^-29kg

So now Energy released would be
(3x10⁸)² * 1.45582 x 10^-29kg
E = 1.310238 x 10^-10J

However something is borthing me, is this the expected energy of the beta particle as well, or is that another separate calculation?

That's energy not accounted for by mass. What form might that energy take?

 

[Gigaz]

That is to a good approximation the expected energy of the Beta particle. But keep in mind that there should actually be a neutrino, and that recoil is also a thing. The neutrino can carry away any part of the energy, whereas the correction for the recoil should be on the order of mass Beta/mass O.

 

[Richie Smash]

I will keep that in mind Gigaz, the exam I'm writing is sort of an introductory to physics in a sense.

That's energy not accounted for by mass. What form might that energy take?

I believe that energy would take the form of heat? and the beta particle will be experiencing like Gigaz said kinetic energy?

 

[PeroK]

I will keep that in mind Gigaz, the exam I'm writing is sort of an introductory to physics in a sense.
I believe that energy would take the form of heat? and the beta particle will be experiencing like Gigaz said kinetic energy?

Heat at an elementary level is kinetic energy of particles.

 

[Richie Smash]

Oh OK, so the energy released is the same as the kinetic energy of the beta particle !

 

[PeroK]

Oh OK, so the energy released is the same as the kinetic energy of the beta particle !

Yes, but look carefully at what the question asked.

 

[Richie Smash]

They ask for the expected energy of the beta particle, and I'm safely assuming that the expected energy differs from the Kinetic energy,

Would it be the potential energy perhaps?

 

[PeroK]

They ask for the expected energy of the beta particle, and I'm safely assuming that the expected energy differs from the Kinetic energy,

Would it be the potential energy perhaps?

Look at your relevant equations.

 

[Richie Smash]

I see that e= mc² is the same as K.E = 1/2mv².

My only guess from this point would simply to be to substitute mass of the beta particle in KG and use the equation.

Then I would subtract it from the kinetic energy I calculated.

 

[PeroK]

I see that e= mc² is the same as K.E = 1/2mv².

My only guess from this point would simply to be to substitute mass of the beta particle in KG and use the equation.

Then I would subtract it from the kinetic energy I calculated.

You don't seem to be entirely on top of the material!

Have you been studying special relativity?

If so, what is the (total) energy of a particle?

 

[Richie Smash]

You don't seem to be entirely on top of the material!

Have you been studying special relativity?

If so, what is the (total) energy of a particle?

Hi, I apologize, I am currently doing revision in order to get better.

I believe the total energy of a particle is the rest energy plus the kinetic energy.

 

[PeroK]

Hi, I apologize, I am currently doing revision in order to get better.

I believe the total energy of a particle is the rest energy plus the kinetic energy.

Yes. So, you need to add the beta particle's rest energy to its kinetic energy.

 

[Richie Smash]

I got it, It would be 0.00055 u * 1.66 x 10-27kg = 9.13 x 10^-30kg

Now I use c² x mass of beta particle = 8x10^-9J

Now add that rest energy and K.E and you will get 8.1 x 10^-8J

 

[PeroK]

I got it, It would be 0.00055 u * 1.66 x 10-27kg = 9.13 x 10^-30kg

Now I use c² x mass of beta particle = 8.217x10^-10J

Now add that rest energy and K.E and you will get 9 x 10^-9J

You need to check your arithmetic.

 

[Richie Smash]

I edited my post.

 

[TSny]

I will do this:
16.99913+0.00055 = 16.99968 u

SO mass deficit = 17.00845-16.99968 = 0.00877 u

This would give the correct mass defect if the numbers 17.00845 u and 16.99913 u were the nuclear masses of N-17 and O-17, respectively. But I believe they are atomic masses. This must be taken into account and it will modify the answer.

 

[Richie Smash]

Hmm, but those are the masses that were given in the question :(

 

[TSny]

Hmm, but those are the masses that were given in the question :(

To see a discussion of dealing with atomic masses in beta decay, see page 12 here http://web.monroecc.edu/manila/webfiles/spiral/7nuclearphysics.pdf

or, for a numerical example, scroll down some in this link http://electron6.phys.utk.edu/phys250/modules/module 5/nuclear_decay.htm