De Broglie wavelength — UPSC Physics Optional CSE 2021 Paper II

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?