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siazzle
@siazzle
January 2021
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Bonjour, pourriez-vous m’aider à répondre à ces questions s’il vous plaît ?
Merci
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scoladan
Verified answer
Bonjour,
a) échographie : onde sonore (onde mécanique)
mobile : onde électromagnétique
b) ΔE = hν
h constante de Planck ≈ 6,63.10⁻³⁴ m².kg.s⁻¹ ou J.s
ν fréquence en Hz
ΔE = |E₀ - E₁| = |-5,14 + 3,03| = 2,11 eV
soit 2,11 x 1,60.10⁻¹⁹ J = 3,376.10⁻¹⁹ J
⇒ ν = ΔE/h
soit ν = 3,376.10⁻¹⁹/6,63.10⁻³⁴ ≈ 5.10¹⁴ Hz
Cette fréquence correspond a une longueur d'onde de :
λ = c/ν = 3.10⁸/5.10¹⁴ ≈ 5,9.10⁻⁷ m
soit environ 590 nm, radiation visible (orange)
1 votes
Thanks 1
siazzle
Merci de votre aide
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Verified answer
Bonjour,a) échographie : onde sonore (onde mécanique)
mobile : onde électromagnétique
b) ΔE = hν
h constante de Planck ≈ 6,63.10⁻³⁴ m².kg.s⁻¹ ou J.s
ν fréquence en Hz
ΔE = |E₀ - E₁| = |-5,14 + 3,03| = 2,11 eV
soit 2,11 x 1,60.10⁻¹⁹ J = 3,376.10⁻¹⁹ J
⇒ ν = ΔE/h
soit ν = 3,376.10⁻¹⁹/6,63.10⁻³⁴ ≈ 5.10¹⁴ Hz
Cette fréquence correspond a une longueur d'onde de :
λ = c/ν = 3.10⁸/5.10¹⁴ ≈ 5,9.10⁻⁷ m
soit environ 590 nm, radiation visible (orange)