91影视

1. The striking of electrons on the molybdenum target produces continuous and characteristic x-rays.
2. The kinetic energy of the incident electrons is 35 keV .
3. Increased energy due to accelerating potential, $鈭�\mathrm{E}=50\mathrm{keV}$

Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the magnetic frequency of the photon and thus, equally, equates to the wavelength of the wave. When the frequency of photons is high, its potential is high.

Formulas:

The kinetic energy gained by the electron is,

$鈭�\mathrm{E}=\mathrm{eV}$ 鈥�.. (1)

Here, e is the charge and V is the accelerating potential difference.

The energy of the photon due to Planck鈥檚 relation is,

$\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{位}}$ 鈥�.. (2)

Here, is the Plank鈥檚 constant, c is the speed of light, and $位$is the wavelength.

Equating both the equations (1) and (2), the value of the minimum wavelength using the data is as follow.

$$\begin{gathered} \mathrm{eV}=\frac{\mathrm{hc}}{{\mathrm{位}}_{\min }} \\ {\mathrm{位}}_{\min }=\frac{\mathrm{hc}}{\mathrm{eV}} \\ \mathrm{eV}=\frac{\mathrm{hc}}{{\mathrm{位}}_{\min }} \\ {\mathrm{位}}_{\min }=\frac{\left(6.63脳{10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(3脳{10}^8\mathrm{m}/\mathrm{s}\right)}{\left(50脳{10}^3\mathrm{keV}\right)\left(1.6脳{10}^{-19}\mathrm{J}/\mathrm{eV}\right)} \\ =24.8脳{10}^{-12}\mathrm{m} \\ =24.8\mathrm{pm} \end{gathered}$$

Hence, the value of the wavelength is 24.8 pm.

The values of wavelengths of the lines $K_{伪}$and $K_{尾}$do not depend on the external potential.

Hence, the wavelengths of the lines remain the same irrespective of the accelerating potential.