Step-by-step Derivation:

1. Electron's Kinetic Energy from de-Broglie Wavelength: The de-Broglie wavelength $\lambda$ of an electron is related to its momentum p by the equation: $\lambda = \frac{h}{p}$ where $h$ is Planck's constant. From this, the momentum of the electron is $p = \frac{h}{\lambda}$.

   The kinetic energy $K$ of the electron with mass $m$ and momentum $p$ is given by: $K = \frac{p^2}{2m}$ Substituting the expression for $p$ from the de-Broglie relation into the kinetic energy equation: $K = \frac{\left(\frac{h}{\lambda}\right)^2}{2m} = \frac{h^2}{2m\lambda^2}$

2. Cut-off Wavelength of Emitted X-rays: In an X-ray tube, when an electron with kinetic energy $K$ strikes the target, it can lose its energy to produce X-ray photons. The cut-off wavelength $\lambda_0$ corresponds to the shortest wavelength (and thus highest energy) X-ray photon that can be emitted. This occurs when the entire kinetic energy of a single electron is converted into the energy of a single photon.

   The energy $E$ of a photon with wavelength $\lambda_0$ is given by the Planck-Einstein relation: $E = \frac{hc}{\lambda_0}$ where $c$ is the speed of light.

   For the cut-off wavelength, the photon energy is equal to the electron's kinetic energy: $K = E \implies K = \frac{hc}{\lambda_0}$

3. Equating Expressions and Solving for $\lambda_0$: We now have two expressions for the kinetic energy $K$ of the electron. By equating them, we can find the relationship between $\lambda$ and $\lambda_0$. $\frac{h^2}{2m\lambda^2} = \frac{hc}{\lambda_0}$ Now, we solve for $\lambda_0$: $\lambda_0 = \frac{hc \cdot (2m\lambda^2)}{h^2}$ Cancel one factor of $h$ from the numerator and denominator: $\lambda_0 = \frac{2mc\lambda^2}{h}$

4. Conclusion: The derived expression for the cut-off wavelength of the emitted X-rays is $\lambda_0 = \frac{2mc\lambda^2}{h}$. This matches option A.

Option Analysis:

• A: $\lambda_{0}=\frac{2 m c \lambda^{2}}{h}$: This matches our derived result. Correct.
• B: $\lambda_{0}=\frac{2 h}{m c}$: Incorrect.
• C: $\lambda_{0}=\frac{2 m^{2} c^{2} \lambda^{3}}{h^{2}}$: Incorrect.
• D: $\lambda_{0}=\lambda$: Incorrect.