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1. The emission spectrum of mercury atoms has a bright green line with wavelength
546.1 nm.
Calculate the frequency of these photons. Show your work.

Respuesta :

Emission spectrum results from the movement of an electron from a higher to a lower energy level. The frequency of the photon is 5.5 * 10^14 Hz.

From the formula;

E = hc/λ

h = Plank's constant =[tex]6.6 * 10^-34[/tex] Js

c = speed of light=  [tex]3 * 10^8[/tex]

λ = wavelength = [tex]546.1 * 10^-9[/tex] m

E =  [tex]6.6 * 10^-34 * 3 * 10^8/546.1 * 10^-9[/tex]

E =[tex]3.63 * 10^-19[/tex] J

Also;

E =hf

Where;

h = Planks's constant

f = frequency of photon

f = E/h

f = [tex]3.63 * 10^-19 J/6.6 * 10^-34[/tex]

f = [tex]5.5 * 10^14[/tex] Hz

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