An X-ray tube operates at 40 kV . Assume that during each...

Initial energy of the electron, $E_0 = 40 \text{ keV}$.

In the first collision, the electron converts $70\%$ of its energy into a photon. The energy of the first photon is:
$E_1 = 0.70 \times 40 \text{ keV} = 28 \text{ keV}$

The remaining kinetic energy of the electron is:
$E_1' = 40 \text{ keV} - 28 \text{ keV} = 12 \text{ keV}$

In the second collision, the electron again converts $70\%$ of its remaining energy into a photon. The energy of the second photon is:
$E_2 = 0.70 \times 12 \text{ keV} = 8.4 \text{ keV}$

The remaining kinetic energy of the electron is:
$E_2' = 12 \text{ keV} - 8.4 \text{ keV} = 3.6 \text{ keV}$

In the third collision, the energy of the third photon is:
$E_3 = 0.70 \times 3.6 \text{ keV} = 2.52 \text{ keV}$

The corresponding wavelengths are given by $\lambda = \dfrac{hc}{E}$. Using $hc \approx 1242 \text{ keV}\cdot\text{pm}$:

$\lambda_1 = \dfrac{1242}{28} \approx 44.3 \text{ pm}$

$\lambda_2 = \dfrac{1242}{8.4} \approx 148 \text{ pm}$

$\lambda_3 = \dfrac{1242}{2.52} \approx 493 \text{ pm}$

These represent the lowest three wavelengths emitted from the tube.