By definition, the deBroglie wavelength is given by:
λ = h p \lambda = \dfrac{h}{p} λ = p h , where h = 6.6 ⋅ 1 0 − 34 J ⋅ s h = 6.6\cdot 10^{-34} J\cdot s h = 6.6 ⋅ 1 0 − 34 J ⋅ s is the Planck constant and p p p is the momentum of the electron. In the relativistic case:
p = m v 1 − v 2 c 2 p = \dfrac{mv}{\sqrt{1-\frac{v^2}{c^2}}} p = 1 − c 2 v 2 m v , where m = 9.1 ⋅ 1 0 − 31 k g m = 9.1\cdot 10^{-31} kg m = 9.1 ⋅ 1 0 − 31 k g is the electon mass, c = 3 ⋅ 1 0 8 c = 3\cdot 10^8 c = 3 ⋅ 1 0 8 speed of light in vacuum and v v v is the electron speed.
Finaly obtain:
λ = h m v 1 − v 2 c 2 \lambda = \dfrac{h}{mv}\sqrt{1-\frac{v^2}{c^2}} λ = m v h 1 − c 2 v 2 .
Substitute numerical values:
a) λ = 6.6 ⋅ 1 0 − 34 9.1 ⋅ 1 0 − 31 ⋅ 1 ⋅ 1 0 8 1 − 1 ⋅ 1 0 16 9 ⋅ 1 0 16 = 6.84 ⋅ 1 0 − 12 m \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 λ = 9.1 ⋅ 1 0 − 31 ⋅ 1 ⋅ 1 0 8 6.6 ⋅ 1 0 − 34 1 − 9 ⋅ 1 0 16 1 ⋅ 1 0 16 = 6.84 ⋅ 1 0 − 12 m
b) λ = 6.6 ⋅ 1 0 − 34 9.1 ⋅ 1 0 − 31 ⋅ 2 ⋅ 1 0 8 1 − 4 ⋅ 1 0 16 9 ⋅ 1 0 16 = 2.7 ⋅ 1 0 − 12 m \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 λ = 9.1 ⋅ 1 0 − 31 ⋅ 2 ⋅ 1 0 8 6.6 ⋅ 1 0 − 34 1 − 9 ⋅ 1 0 16 4 ⋅ 1 0 16 = 2.7 ⋅ 1 0 − 12 m
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