Answer to Question #135157 in Quantum Mechanics for Bhupinder

Question

An electron is accelerated through potential of 150 Volts, wavelength associated with it is ?

Answer 1

We may calculate de Broglie wavelength as

$\lambda = \dfrac{h}{p} = \dfrac{h}{mv}.$ The kinetic energy is, on the one hand, $E_k = \dfrac{mv^2}{2}$ and, on the other hand, $E_k = q_e \Delta\varphi.$

Therefore, $\lambda =\dfrac{h}{mv} = \dfrac{h}{\sqrt{2q_em\Delta\varphi}} = \dfrac{6.6\cdot10^{-34}\,\mathrm{J\cdot s}}{\sqrt{2\cdot 9.1\cdot10^{-31}\,\mathrm{kg}\cdot1.6\cdot10^{-19}\,\mathrm{C}\cdot 150\,\mathrm{V}}} = 10^{-10}\,\mathrm{m} = 1$ Å.