**Calculate the energy in J/photon:**

$\mathbf{E}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{77}\mathbf{\times}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\frac{\overline{)\mathbf{kJ}}}{\overline{)\mathbf{mol}}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{J}}{\mathbf{1}\mathbf{}\overline{)\mathbf{kJ}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}}}{\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times}{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{atoms}}$

**E = 1.29× 10 ^{–15} J/atom = 1.29× 10^{–15} J/photon**

**Calculate wavelength:**

$\overline{){\mathbf{E}}{\mathbf{=}}{\mathbf{hv}}}{\mathbf{}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

$\overline{){\mathbf{\lambda}}{\mathbf{=}}\frac{\mathbf{c}}{\mathbf{v}}}\mathbf{}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{\mathbf{hc}}{\mathbf{\lambda}}}$

$\mathbf{\lambda}\mathbf{=}\frac{\mathbf{hc}}{\mathbf{E}}$

X-ray diffractometers often use metals that have had their core electrons excited as a source of X rays. Consider the 2*p *→ 1*s* transition for copper, which is called the *Kα* transition. Calculate the wavelength of X rays (in angstroms) given off by the *Kα* transition if the energy given off by a mole of copper atoms is 7.77 × 10^{5} kJ . (Note that 1Å = 10^{-10} m)

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