Photon giving energy ?

[Dino]

Dear reader of this topic, I was wondering what would happen if a photon gives his energy to a proton or a neutron in a atom. (Imagine that a photon would come across a single atom in space for example and he would miss the elektrons)

Would the balance of the protons and neutrons disturb and what would happen next ?

(Please correct me if I am saying anything wrong)

 

[Strange]

There is a good summary of the ways a gamma photon can interact with an atom here: http://www.physics.rutgers.edu/ugrad/205/manuals/matter.pdf

(pages 5 & 6)

 

I don't think lower energy photons will interact with the nucleus at all.

[swansont]

The balance of neutrons and protons will affect whether the photon is absorbed, since that dictates the energy level structure of the nucleus. So you could cause a nuclear excitation. It is possible that a photon of sufficient energy could eject a neutron or proton.

[Dino]

So does that mean that if a photon with sufficient energy to eject a proton of a element that it would change into another element ?

So for example: From Helium to Hydrogen ?

Is this correct ?

[swansont]

So does that mean that if a photon with sufficient energy to eject a proton of a element that it would change into another element ?

So for example: From Helium to Hydrogen ?

Is this correct ?

Yes. Or change isotope, if a neutron were ejected.

[Dino]

So does that mean that if technology is able to create a photon with sufficient energy and to let him move in the right direction to hit a proton we could change for example Helium into Hydrogen, this would be a great way to "change" all sorts of elements without the fear of extinction of them on this planet.

Edited May 27, 2017 by Dino

[swansont]

So does that mean that if technology is able to create a photon with sufficient energy and to let him move in the right direction to hit a proton we could change for example Helium into Hydrogen, this would be a great way to "change" all sorts of elements without the fear of extinction of them on this planet.

Energy-wise this is a very inefficient way of transmuting elements.

[Dino]

So for example: a Helium-4 isotope to a Hydrogen-3 isotope taking 1 proton away would take if my calculations are correct:

Helium-4 has 2 protons and 2 neutrons,

2 * 1,007276466812 = 2,014552934 u and 2 * 1,00866491600 = 2,017329832

2,014552934 + 2,017329832 = 4,031882766 u

Helium - 4 mass = 4,002603

4,031882766 - 4,002603 = 0,029279766 u

1 u = 931,494061 MeV

0,02927976 * 931,494061 = 27,27392814 MeV

27,27392814 / 4 = 6,818482034 MeV

So does that mean that if I have calculated it correctly it takes 6,818482034 MeV to remove a proton of the nucleus from a Helium-4 isotope ?

Edited May 27, 2017 by Dino

[Sensei]

So for example: a Helium-4 isotope to a Hydrogen-3 isotope taking 1 proton away would take if my calculations are correct:

Helium-4 has 2 protons and 2 neutrons,

2 * 1,007276466812 = 2,014552934 u and 2 * 1,00866491600 = 2,017329832

2,014552934 + 2,017329832 = 4,031882766 u

Helium - 4 mass = 4,002603

4,031882766 - 4,002603 = 0,029279766 u

1 u = 931,494061 MeV

0,02927976 * 931,494061 = 27,27392814 MeV

27,27392814 / 4 = 6,818482034 MeV

So does that mean that if I have calculated it correctly it takes 6,818482034 MeV to remove a proton of the nucleus of a Helium-4 isotope ?

 

Not really.

 

[math]^4_2He + 19.82 MeV \rightarrow ^3_1H + p^+[/math]

 

Take mass of Helium-4 4.0026 u

multiply by 931.494 MeV/u

subtract 2 electrons 0.511 MeV each,

you will receive 3727.38 MeV

 

Repeat the same with Hydrogen-3 3.01605 u

you will receive 2808.92 MeV.

 

3727.38 MeV + 19.82 MeV = 2808.92 MeV + 938.272 MeV

(without taking into account momentum)

 

Helium-4 has one of the largest energies needed to eject proton or neutron, from the all isotopes.

Edited May 27, 2017 by Sensei

[Dino]

So for example: to go from Fluorine to Oxygen-18 isotope,

 

 

Fluorine = 18,99840 u

18,99840 * 931,494 = 17696,89561 MeV

17696,89561 - (9 * 0,511) = 17692,30 MeV

 

Oxygen-18 isotope = 17,99916 u

17,99916 * 931,494 = 16766,11 MeV

16766,11 - (8 * 0,511) = 16762,02 MeV

 

17692,30 MeV + ? = 16762,02 MeV + 938,272 MeV

17692,30 MeV + 7,99 MeV = 16762,02 MeV + 938,272 MeV

 

Are these calculations corrrect ?

Edited May 28, 2017 by Dino

[Sensei]

Are these calculations corrrect ?

 

Yes. My app is showing 7.99485 MeV as well.

[Dino]

So to eject a proton from a Fluorine nucleus there is a photon needed with 7,99485 MeV of energy going in the right direction to get absorbed by a proton of a Fluorine nucleus ?

To calculate the wavelength of the photon there is this equation:

[latex]

{E}=\frac{\ hc}{\lambda }

[/latex]

or

[latex]

{\lambda}=\frac{\ hc}{E }

[/latex]

 

[latex]

{\lambda(\mu m)}=\frac{\ 1,24}{7994850 }

[/latex]

 

which brings up to a wavelenght of 1,550998455 * 10^-07 micrometer,

multiplied by a 1000 to give nanometers it is 1,550998455 * 10^-04 nanometer

which is 0,0001550998455 nanometer.

 

Is this correct ?

 

(Please correct me if I am saying anthing wrong)

Edited May 28, 2017 by Dino