Green light has a wavelength of about 550 nm. Through what potential difference must an electron be accelerated to have this wavelength?
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11
QV = (KE)electron = (mv2)/2 = p2/2m
since p = mv
But lambda = h/p
therefore, p = h/(lambda)
so that, V = p2/(2mQ) = h2/(2mQ(lambda)2) = (6.6 x 10-34)2/(2 x 9.1 x 10-31 x 1.6 x 10-19 x (550 x 10-9)2)
V = 4.9 x 10-6 Volts
since p = mv
But lambda = h/p
therefore, p = h/(lambda)
so that, V = p2/(2mQ) = h2/(2mQ(lambda)2) = (6.6 x 10-34)2/(2 x 9.1 x 10-31 x 1.6 x 10-19 x (550 x 10-9)2)
V = 4.9 x 10-6 Volts
Answered by
4
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 . 16. 10^–14 = 1.87 . 10^–6
V = 1.9 µV
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 . 16. 10^–14 = 1.87 . 10^–6
V = 1.9 µV
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