For tungsten, the wavelength of the K_α X-ray is 21.3 pm . Ejecting...

The energy of the emitted $K_\alpha$ X-ray photon is given by:

$E_{K_\alpha} = \dfrac{hc}{\lambda}$

Using $hc \approx 1240 \text{ keV pm}$ and $\lambda = 21.3 \text{ pm}$:

$E_{K_\alpha} = \dfrac{1240 \text{ keV pm}}{21.3 \text{ pm}} \approx 58.2 \text{ keV}$

The $K_\alpha$ X-ray is emitted when an electron transitions from the L shell to a vacancy in the K shell. Therefore, the energy difference between the K and L shells is:

$E_{K_\alpha} = E_K - E_L$

Given the binding energy of the L shell $E_L = 11.3 \text{ keV}$, the binding energy of the K shell is:

$E_K = E_{K_\alpha} + E_L = 58.2 \text{ keV} + 11.3 \text{ keV} = 69.5 \text{ keV}$

To permit the emission of $K_\alpha$ X-rays, the incident electrons from the cathode must have sufficient kinetic energy to knock out an electron from the K shell of the tungsten target. Thus, the minimum kinetic energy required is $69.5 \text{ keV}$.

The minimum accelerating voltage $V$ across the X-ray tube is:

$V = \dfrac{E_K}{e} = 69.5 \text{ kV}$