Question #135157
An electron is accelerated through potential of 150 Volts, wavelength associated with it is ?
1
Expert's answer
2020-09-28T08:08:43-0400

We may calculate de Broglie wavelength as

λ=hp=hmv.\lambda = \dfrac{h}{p} = \dfrac{h}{mv}. The kinetic energy is, on the one hand, Ek=mv22E_k = \dfrac{mv^2}{2} and, on the other hand, Ek=qeΔφ.E_k = q_e \Delta\varphi.

Therefore, λ=hmv=h2qemΔφ=6.61034Js29.11031kg1.61019C150V=1010m=1\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.


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