Kerala KEAM 5 June 2024 Solved Paper

The de Broglie wavelength $\lambda$ associated with a particle (in this case, an electron) is given by the formula:$$\lambda = \frac{h}{p}$$where:- $h$ is Planck's constant ($6.626 \times 10^{-34} \, J·s$),- $p$ is the momentum of the particle.The momentum $p$ of an electron accelerated by a potential $V$ is:$$p = \sqrt{2 m_e e V}$$where:- $m_e$ is the mass of the electron ($9.11 \times 10^{-31} \, kg$),- $e$ is the charge of the electron ($1.6 \times 10^{-19} \, C$),- $V$ is the potential difference (in this case, 81 V).Substituting these values into the formula for $p$:$$p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19} \times 81}$$Now, the de Broglie wavelength $\lambda$ can be calculated using the above formula. For this case, the wavelength $\lambda$ will be in the range of X-rays.Therefore, the correct answer is:$$\boxed{D) X-rays}$$ Quick Tip: The de Broglie wavelength of a particle depends on its momentum. For high-energy electrons, such as those accelerated by 81 V, the wavelength lies in the X-ray region of the electromagnetic spectrum.