Calculating the Recoil Energy of 57Co Emitting a 14.4 keV Gamma Ray

[dranger35]

What is the recoil energy (in eV) of a 57Co nucleus
emitting a 14.4 keV gamma ray?

If anyone knows the equation that I can use, thanks.

 

[dextercioby]

Since the energy of the photon is small (i think it's not [itex] \gamma[/itex],are u sure it's not MeV...?),you can use the nonrelativistic formula for KE...

For a nice use of the total momentum conservation law ,assume the nucleus to be initially at rest.

Daniel.

 

[thepatientmental]

The recoil energy of a nucleus can be calculated using the equation E_recoil = (E_gamma)^2/2mc^2, where E_gamma is the energy of the gamma ray and m is the mass of the nucleus. In this case, the energy of the gamma ray is given as 14.4 keV, which is equivalent to 1.44 x 10^-5 joules.

The mass of a 57Co nucleus is approximately 9.474 x 10^-26 kg. Plugging in these values into the equation, we get E_recoil = (1.44 x 10^-5)^2/2(9.474 x 10^-26)(3 x 10^8)^2 = 2.592 x 10^-19 joules.

Converting this to electron volts (eV), we have 2.592 x 10^-19 joules x (6.24 x 10^18 eV/joule) = 1.616 eV. Therefore, the recoil energy of a 57Co nucleus emitting a 14.4 keV gamma ray is approximately 1.616 eV.