De Broglie wavelength
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De Broglie wavelength

[From: ] [author: ] [Date: 11-08-18] [Hit: ]
V =h^2 / 2 .m . e . λ^2 =44.0 . 10^–68/ 2 .......
Through what potential difference must an electron be accelerated from rest to have a de Broglie wavelength of 600 nm?

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λ = h/(m * v) ----> m * v = p = √(2m * KE) = √(2m * e * V)

--> λ = h/(√(2m * e * V))

--> V = h^2/(2m * e * V * λ^2)
= (6.63 x 10^-34)^2/(2 * 9.11 x 10^-31 * 1.6 x 10^-19 * (6 x 10^-7)^2)
= 4.19 x 10^-6 volts

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λ = h/mv

mv = h / λ

KE = (mv)^2 / 2 m = h^2 / 2m . λ^2 = Ve

V = h^2 / 2 .m . e . λ^2 = 44.0 . 10^–68 / 2 . 9.1 . 10^–31 . 1.6 . 10^–19 . 36 . 10^–14 = 4.2 . 10^–6

V = 4.2 µV

very small for a voltage . Please check that I've done the numbers correctly. The equations look ok
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keywords: wavelength,De,Broglie,De Broglie wavelength
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