Answer
$2.16\times10^{-11}m$, $1.84\times10^{-11}m$.
Work Step by Step
The energies of the emitted x rays are equal to the energy differences between the shells.
A $K_{\alpha}$ x ray is produced in a transition from L to K.
$$\Delta E = E_L-E_K=-12000eV-(-69500eV)=57500eV$$
From its energy, we can calculate the wavelength of the x ray.
$$\lambda =\frac{hc}{E}=\frac{1.24\times10^{-6}eV\cdot m}{57500eV }=2.16\times10^{-11}m$$
A $K_{\beta}$ x ray is produced in a transition from M to K.
$$\Delta E = E_M-E_K=-2200eV-(-69500eV)=67300eV$$
From its energy, we can calculate the wavelength of the x ray.
$$\lambda =\frac{hc}{E}=\frac{1.24\times10^{-6}eV\cdot m}{67300eV }=1.84\times10^{-11}m$$
