For a cartain metal, the `K` obsorption edge is at `0.72 Å`. The wavelength of `K_(alpha), K_(beta). And K_(gamma)` lines of of `K` series are `0.210 Å , 0.192 Å, and 0.180 Å, respectively. The eneggies of `K, L and M` orbit are `E_(K), E_(L) and E_(M)`, respectively. Then

Options

A. `E_(K)=-13.40eV`
B. `E_(L)=-7.52KeV`
C. `E_(M)=-3.21KeV`
D. `E_(K)=13.04KeV`

Answer

A, B, C

Explanation

Energy of `K` absorption edge
` E'_(K)=(1242eVnm)/(0.0172nm)=72.21xx10^(3)eV=72.21KeV`
Energy of `K_(Kalpha)` line `E_(Kalpha)=(hc)/(elambda_(alpha))`
`(1242eVnim)/(0.021nm)=59.14KeV`
Energy of `K` shell `=(K_(Kalpha)-E'_(K))`
`=(59.14-72.21)KeV=-13.04keV`
Energy of `L` shell `=E_(kbeta)-72.21keV`
`=64.69keV-72.21keV=-7.52keV`
Energy of M shell =`K_(kgamma)-E_(k)=(1242eVnm)/(0.018nm)-72.21keV`
` =69keV-72.21keV=-3.21keV`