The energy of the n=2 state in a given element is `E_2 =-2870 eV`. Given that the wavelengths of lthe `K_alpha` and `K_beta` lines are 0.71 nm and 0.63 nm respectively, determine the energies `E_1` and `E_3`.

Answer

A, B, C, D

Explanation

`lambda_(K_alpha) = 0.71 nm - 7.1 Å`
`E_2 - E_1 = (127375)/(7.1) = 1743 eV`
`:. E_1 = E_2 - 1743`
=-2870 - 1743
=- 4613 eV
`lambda_(K_beta) = 0.63 nm`
`=6.3 Å`
`E_3 -E_1 = (12375)/(6.3)`
(1964) eV
`:. E_3 = E_1+ 1964`
=- 4613 +1946
=- 2649 eV.