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:

or

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)

 

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:

or

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 ?

I ONLY NEED THE ANSWER IF THE FORMULA IS CORRECT !!!!

SORRY DID NOT SEE FOR HOMEWORK (NO ANSWERS)

 

(Please correct me if I am saying anthing wrong)

 

Edited June 1, 20178 yr by Dino

Are you sure you're using consistent units for the various quantities? That is, mks units for everything, or similarly consistent values in some other system? If your speed of light is in meters per second and your h in Joule-seconds and your wavelength in meters, then your energy should be in Joules, I believe.

Are you sure you're using consistent units for the various quantities? That is, mks units for everything, or similarly consistent values in some other system? If your speed of light is in meters per second and your h in Joule-seconds and your wavelength in meters, then your energy should be in Joules, I believe.

He used (sort of) Planck constant in electro volts * seconds, and energy in electron volts. But multiplied them by 10^6.

 

[math]E=\frac{h c}{\lambda} = \frac{4.135667*10^{-15} [eV*s] * 299792458 [\frac{m}{s}]}{\lambda [m]} = \frac{1.24*10^{-6}[eV*m]}{\lambda [m]}[/math]

 

[math]7994850 [eV] = \frac{1.24*10^{-6} [eV*m]}{\lambda [m]}[/math]

 

[math]\lambda [m]= \frac{1.24*10^{-6} [eV*m]}{7994850 [eV]} = 1.55*10^{-13} [m] = 0.155 [pm] = 155 [fm][/math]

 

 

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 induce decay of Flourine-19 there can be used much less energy.

 

[math]^{19}_9F + 4.01374 MeV \rightarrow ^{15}_7N + ^4_2He[/math]

Edited June 2, 20178 yr by Sensei