What is the wavelenth of the `0.186 MeV` gamma- ray photon emitted by radium `._(88)^(226) Ra`?

Explanation

The photon energy is the difference between two nuclear energy levels. Equation `E_(i )- E_(f) =hf` gives the relation between the energy level separation `Delta E` and the frequency f of the photon as `Delta E =hf` . Since `f lambda =c` , the wavelength of the photon is `lambda -hc//Delta E`.
First, we must convert the photon energy into joule:
`Delta E =(0.186 xx 10^(6) eV)((1.60 xx 10^(-19))/1 eV)`
` = 2.98 xx 10^(-14) J`
The wavelength of the photon is
`lambda=(hc)/(Delta) E =((6.63 xx 10^(-34) J.s)(3.00 xx 10^(8) m s^(-1)))/(2.98 xx 10^(-14)J)`
`=6.67+10^(-12)m`.