Question

X-rays are produced when the electron stream in an x-ray tube knocks an electron out of a low-lying shell of an atom in the target, and an electron from a higher shell falls into the lower-lying shell. The x-ray is the photon given off as the electron falls into the lower shell. The most intense x-rays produced with a copper target have wavelenghts of 1.542A and 1.392A. These x-rays are produced when an electron from the L or M shell falls into the K shell of teh copper atom. Calculate the energy separation in eV of the K,L, and M shells in copper.

Answer 1

The electrons lie in K, L and M shells. More intense x-rays would be emitted with wavelengths 1.542 and 1.392 A.

Energy = hc/

h = 6.63*10^-34 m² kg/s

c = 3*10⁸ m/s

= 1.542 = 1.542*10^-10 m

E₁ = (6.63*10^-34)(3*10⁸)/(1.542*10^-10) = 12.90*10^-16 J

= 1.392 = 1.392*10^-10 m

E₂ = (6.63*10^-34)(3*10⁸)/(1.392*10^-10) = 14.29*10^-16 J

The E1 < E2, so E₂ is the energy separation for K and M shell and E₁ is the separation for K and L shell.

To calculate separation between L and M shell, (14.29*10^-16 J)-(12.90*10^-16) = 1.39*10^-16 J

The energy separation between K and L is 12.90*10^-16 J = 7998 eV

The energy separation between L and M is 1.39*10^-16 J = 862 eV

The energy separation between K and M is 14.29*10^-16 J = 8860 eV