If Bohr's 3rd orbit is circumference. Find out of the de broglie electronic wave length .
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heya..☺️☺️⬅️☢
According to de Broglie, the circumference of a stationary orbit must be an integral number of wavelengths. nλ=2πr. Recall that for a hydrogen
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Radius of 3rd orbit = r = R(3)^2
= 9 R
[Here R = Radius of first orbit of hydrogen atom]
mvr = nh / (2 π)
h/(mv) = 2πr / n
λ = h / (mv) = 2π × 9R / 3
λ = 6πR
λ = 6 π × (0.529 Å)
λ = 9.97 Å
De-Broglie wavelength is 9.97 Å
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