cse • paper_2 • quantum_mechanics
[UPSC CSE 2021] Foundations of Quantum Mechanics (Q15, 15M)
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?
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