Animated Solution for Physics - Dual Nature of Matter and Radiation: Electrons with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is

Imagine an electron zooming towards a metal target inside an X-ray tube. This electron isn't just a particle; it behaves like a wave with a de-Broglie wavelength $\lambda$. When it slams into the target, it rapidly decelerates, and its kinetic energy is converted into an X-ray photon.

Our goal is to find the cut-off wavelength (${\lambda }_0$) of the emitted X-rays. The cut-off wavelength represents the absolute shortest wavelength possible, which corresponds to the maximum possible energy of the emitted photon.

First, we need to determine the kinetic energy of the incoming electron. According to de-Broglie's relation, the momentum $p$ of the electron is given by:

Now, we relate this momentum to the kinetic energy $K$ of the electron. Using the standard classical mechanics formula $K=\frac{p^2}{2m}$, we substitute our expression for momentum:

This is the total kinetic energy the electron brings to the collision.

To get the cut-off wavelength ${\lambda }_0$, we must assume the most extreme case: the electron transfers 100% of its kinetic energy into a single X-ray photon. The energy of this photon is given by Planck's equation $E=\frac{hc}{{\lambda }_0}$.

Equating the photon's energy to the electron's kinetic energy, we get:

Now, it's just a matter of simple algebra. We can cancel one $h$ from both sides:

Rearranging the terms to isolate ${\lambda }_0$, we arrive at our final expression:

This perfectly matches option (a). It's a beautiful demonstration of how matter waves and electromagnetic waves are interconnected through energy conservation!