Question #114899

Find the deBroglie wave length of (a)an electron whose speed is1.0×108m/s,and (b)an electron whose speed is2.0×108m/s.


1
Expert's answer
2020-05-11T20:06:20-0400

By definition, the deBroglie wavelength is given by:

λ=hp\lambda = \dfrac{h}{p} , where h=6.61034Jsh = 6.6\cdot 10^{-34} J\cdot s is the Planck constant and pp is the momentum of the electron. In the relativistic case:

p=mv1v2c2p = \dfrac{mv}{\sqrt{1-\frac{v^2}{c^2}}} , where m=9.11031kgm = 9.1\cdot 10^{-31} kg is the electon mass, c=3108c = 3\cdot 10^8 speed of light in vacuum and vv is the electron speed.

Finaly obtain:

λ=hmv1v2c2\lambda = \dfrac{h}{mv}\sqrt{1-\frac{v^2}{c^2}} .

Substitute numerical values:

a) λ=6.610349.11031110811101691016=6.841012m\lambda = \dfrac{6.6\cdot 10^{-34}}{9.1\cdot 10^{-31}\cdot 1\cdot10^8}\sqrt{1-\frac{1\cdot10^{16}}{9\cdot 10^{16}}} = 6.84\cdot 10^{-12} m

b) λ=6.610349.11031210814101691016=2.71012m\lambda = \dfrac{6.6\cdot 10^{-34}}{9.1\cdot 10^{-31}\cdot 2\cdot10^8}\sqrt{1-\frac{4\cdot10^{16}}{9\cdot 10^{16}}} = 2.7\cdot 10^{-12} m


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