AIIMS 2018 Physics de Broglie Wavelength MCQ Question

The de Broglie wavelength $\lambda$ is derived from the relationship between a particle's momentum and its wave-like properties, expressed as $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is the momentum. At the highest point of its trajectory, the particle's momentum can be expressed as $p = mu \sin \theta$, leading to the formula $\lambda = \frac{h}{mu \sin \theta}$. Other options are not applicable as they do not provide relevant information or alternative expressions for the de Broglie wavelength in this context. Understanding this concept is crucial for grasping the wave-particle duality of matter in quantum mechanics.