# What is the Mass of a Deuteron?

• danielatha4
In summary, a proton and a neutron combine to form a deuteron, emitting a gamma ray with an energy of 2.39 MeV. The mass of the deuteron is calculated to be 3.34318*10^-27 kg, taking into account the rest energy and the energy of the photon. The binding energy between the proton and neutron is negative, leading to a coupling distance of 0.746 fm. This simple magnetic dipole binding does not require the use of extra-nucleonic strong forces.
danielatha4

## Homework Statement

A proton (1.6726×10-27 kg) and a neutron (1.6749×10-27 kg) at rest combine to form a deuteron, the nucleus of deuterium or "heavy hydrogen". In this process, a gamma ray (high-energy photon) is emitted, and its energy is measured to be 2.39 MeV (2.39×106 eV). Keeping all five significant figures, what is the mass of the deuteron? (Assume you can neglect the small kinetic energy of the recoiling deuteron.)

## Homework Equations

Rest Energy = m c^2
Ebefore=Eafter

## The Attempt at a Solution

I found the energy before the combining by

c^2 (1.6726×10-27+1.6749×10-27) = 3.0086*10^-10 J

this should equal the energy of the deuteron (mc^2) + the energy of the photon

energy of the photon in joules: 3.8292*10^-13 J

therefore:

3.0086*10^-10 = mdc^2 + 3.8292*10^-13

md=3.3433*10^-27 kg

However, this isn't right

Do you know the correct answer? You should be right, unless there is a rounding error, but I know that the mass of deuteron is just about what you've stated. Make sure you're giving the answer in the correct units (usually we talk about particle masses in terms of rest mass and give the units in MeV).

danielatha4 said:

## The Attempt at a Solution

I found the energy before the combining by

c^2 (1.6726×10-27+1.6749×10-27) = 3.0086*10^-10 J

this should equal the energy of the deuteron (mc^2) + the energy of the photon

energy of the photon in joules: 3.8292*10^-13 J

therefore:

3.0086*10^-10 = mdc^2 + 3.8292*10^-13

md=3.3433*10^-27 kg

However, this isn't right

you're right, it's wrong- binding energy is negative so the above equation should read:

3.0086*10^-10 = mdc^2 - 3.8292*10^-13,

so

md=3.34318*10^-27 kg

only a fraction out.

What gets me is that, where Ed is the deuteron binding in joules, Up and Un are the proton and neutron magnetions respectivey and Uo is the magnetic permeability of free space:

(UpUnUo/Ed)^1/3 = 0.746 fm

in other words a coupling distance between proton and neutron forming a deuteron of about half a nuclear radius from simple magnetic dipole binding, without having to invoke an extra-nucleonic strong force at all. Protons will also bind nicely once the magnetic attraction overcomes electrostatic repulsion at about 1.8 fm, at an energy of about 532Kev. Of course there are no quantum numbers involved here so the digits are subject to change, but not the scale, which is one of nuclear interactions.

E.T. said:
you're right, it's wrong- binding energy is negative so the above equation should read:

3.0086*10^-10 = mdc^2 - 3.8292*10^-13,

so

md=3.34318*10^-27 kg

only a fraction out.
Actually, you have it backwards.

I know what you happened. You got all the steps right, but you probably need to use the speed of light as 3*10^9 in the program. I did this same problem for my class and i figured it out by your post since it didn't except my answer wither until i did that.

E.T. said:
in other words a coupling distance between proton and neutron forming a deuteron of about half a nuclear radius from simple magnetic dipole binding, without having to invoke an extra-nucleonic strong force at all. Protons will also bind nicely once the magnetic attraction overcomes electrostatic repulsion at about 1.8 fm, at an energy of about 532Kev. Of course there are no quantum numbers involved here so the digits are subject to change, but not the scale, which is one of nuclear interactions.

although i now realize the formula is wrong by a factor of 4pi, and the separation distance is also wrong- I have been looking at Yukawa potentials, but formalism constantly gets in the way of a formula which is actually useul when it comes to replicating numbers on a calculator. It's nice to know people are out there to help and don't just ignore wrongheaded ideas in favour of stating the equally incorrect (for example that the deuteron is heavier than its constituents), or repeating indeciperable jargon. :| Einstein would be ashamed!

## What is a deuteron?

A deuteron is a type of atomic nucleus that contains one proton and one neutron. It is the nucleus of the deuterium isotope of hydrogen.

## What is the energy of a deuteron?

The energy of a deuteron is the total amount of energy contained within the nucleus. This includes the energy of the protons and neutrons, as well as any potential energy from their interactions.

## How is the energy of a deuteron calculated?

The energy of a deuteron is calculated using Einstein's famous equation, E=mc². This equation relates the mass of an object to its energy, and can be used to determine the energy of a deuteron based on its mass.

## Why is the energy of a deuteron important?

The energy of a deuteron is important because it is a key factor in understanding nuclear reactions and processes. It also plays a role in understanding the stability and behavior of atoms and molecules.

## Can the energy of a deuteron be changed?

Yes, the energy of a deuteron can be changed through various nuclear reactions such as fusion or fission. It can also be affected by external factors such as temperature and pressure.

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