Question

When electrons bombard a molybdenum target, they produce both continuous and characteristic x-rays as shown in Fig. 40-13. In that figure the kinetic energy of the incident electrons is 35.0 keV. If the accelerating potential is increased to 50.0 keV, (a) what is the value of ${\mathbf{\lambda }}_{\mathbf{min}}$, and (b) do the wavelengths of the role="math" localid="1661497027757" ${\mathbf{k}}_{\mathbf{\alpha }}$and ${\mathbf{k}}_{\mathbf{\beta }}$lines increase, decrease, or remain the same?

Short answer

1. The value of the minimum wavelength is 24.8 pm .
2. The wavelengths of the lines${\mathrm{k}}_{\mathrm{\alpha }}$ and ${\mathrm{k}}_{\mathrm{\beta }}$remain same.

Step-by-step solution

The given data:

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, $∆\mathrm{E}=50\mathrm{keV}$

Understanding the concept of wavelength due to accelerating potential:

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,

$∆\mathrm{E}=\mathrm{eV}$ ….. (1)

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

The energy of the photon due to Planck’s relation is,

$\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ ….. (2)

Here, is the Plank’s constant, c is the speed of light, and $\lambda$is the wavelength.

 Step 3: (a) Calculation of the minimum wavelength:

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

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

Hence, the value of the wavelength is 24.8 pm.

(b) Calculation of the wavelengths of Kα and Kβ lines:

The values of wavelengths of the lines $K_{\alpha }$and $K_{\beta }$do not depend on the external potential.

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