cse • paper_2 • quantum_mechanicscse-2021-subject-05-003
De Broglie wavelength — UPSC CSE 2021 Physics PYQ
Buy PYQ Solution Manual →A particle of rest mass m_0 has a kinetic energy K, show that its de Broglie wavelength is given by \lambda=\frac{hc}{\sqrt{\left[K\left(K+2m_0c^2\right)\right]}}. Hence calculate the wavelength of an electron of kinetic energy 2\mathrm{MeV}. What will be the value of \lambda if K << m_0c^2?
Discussion (0)
Sign in to post a solution, derivation, or discussion.
No comments yet. Start the conversation!