Through what potential difference must an electron be accelerated from rest to have a de Broglie wavelength of 600 nm?
-
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
--> 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
-
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
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