Find the wavelength (in nanometers) of a yellow caution
lightwhose frequency is 5.4 x 1014 s-1.

λ = 555.55 nm

Explanation:

the photon energy equation

∴ h = 6.626 E-34 J*sPlanck's constant

∴ f = 5.4 E14 s-1

⇒ E = (6.626 E-34 J*s)(5.4 E14 s-1) = 3.578 E-19 J

∴ E = hc/λ

⇒ λ = hc/E

∴ c ≅ 3.0 E8 m/s

⇒ λ = ((6.626 E-34 J*s)(3.0 E8 m/s))/(3.578 E-19 J)

⇒ λ = (5.555 E-7 m)×(1 E9 nm/m) = 555.55 nm

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