Expected energy of a beta particle

Click For Summary

Homework Help Overview

The discussion revolves around the expected energy of a beta particle emitted during the decay of Nitrogen-17 into an isotope of Oxygen. Participants are analyzing the provided mass values and the implications of conservation of mass and energy in the context of beta decay.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the given masses and the calculation of energy using E=mc². Questions arise regarding the necessity of all mass values provided and whether they contribute to the calculation of kinetic energy or other forms of energy.

Discussion Status

The discussion is active, with participants sharing their calculations and questioning the relevance of the mass values. Some guidance has been offered regarding the interpretation of energy forms and the role of the neutrino in the decay process. There is an ongoing exploration of how to accurately account for the energy associated with the beta particle.

Contextual Notes

There is a noted concern about whether the masses provided are atomic or nuclear masses, which could affect the calculations. Participants are also considering the implications of the neutrino's presence and its impact on the energy distribution in the decay process.

Richie Smash
Messages
293
Reaction score
15

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...
 
Physics news on Phys.org
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.
 
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?
 
Richie Smash said:
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.
 
  • Like
Likes   Reactions: Richie Smash
Richie Smash said:

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

The mass deficit?
 
Richie Smash said:
The mass deficit?

Exactly.
 
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?
 
Richie Smash said:
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?
 
  • #10
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.
 
  • Like
Likes   Reactions: Richie Smash
  • #11
I will keep that in mind Gigaz, the exam I'm writing is sort of an introductory to physics in a sense.

PeroK said:
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?
 
  • #12
Richie Smash said:
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.
 
  • #13
Oh OK, so the energy released is the same as the kinetic energy of the beta particle !
 
  • #14
Richie Smash said:
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.
 
  • #15
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?
 
  • #16
Richie Smash said:
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.
 
  • #17
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.
 
  • #18
Richie Smash said:
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?
 
  • #19
PeroK said:
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.
 
  • #20
Richie Smash said:
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.
 
  • #21
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
 
Last edited:
  • #22
Richie Smash said:
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.
 
  • #23
I edited my post.
 
  • #24
Richie Smash said:
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.
 
  • Like
Likes   Reactions: PeroK
  • #25
Hmm, but those are the masses that were given in the question :(
 
  • #26
  • Like
Likes   Reactions: Richie Smash

Similar threads

How to calculate gamma ray energies following beta decay?
  • · Replies 12 ·
Replies
12
Views
2K
How can I calculate the rate of energy absorption from nuclear decay?
  • · Replies 23 ·
Replies
23
Views
3K
Bombardment of Nitrogen-14 with an alpha particle
  • · Replies 2 ·
Replies
2
Views
5K
Energy Loss in Plastic Scintillator
  • · Replies 8 ·
Replies
8
Views
2K
What Was the Initial Energy of a Positron Entering a Plastic Slab?
  • · Replies 10 ·
Replies
10
Views
3K
Using energy considerations to analyse particle motion
  • · Replies 3 ·
Replies
3
Views
1K
Determining beta as a function of relativistic momentum
  • · Replies 1 ·
Replies
1
Views
3K
Nuclear Physics help please (alpha decay of a Po-216 atom)
  • · Replies 1 ·
Replies
1
Views
3K
Invariant mass and energy balance
  • · Replies 4 ·
Replies
4
Views
2K
Heat Capacity of Distinguishable Particles
  • · Replies 2 ·
Replies
2
Views
2K