wavelength increases, energy 3. As frequency increases, energy increoses re 4. T

Question

wavelength increases, energy 3. As frequency increases, energy increoses re 4. The intense Lyman- hydrogen solar emission line at 121.58 nm is strongly absorbed by O, in the stratosphere. This results in O,'s dissociation and the ultimate formation of ozone, O,, to form the ozone layer. Calculate the energy of one photon with a wavelength of 121.58nm. a. b. Calculate the energy for a mole of photons of wavelength 121.58 nm. (Remember that 1 mole = 6.022 × 1023) The Lyman- hydrogen solar emission line at 121.58 nm corresponds to an electron dropping from a higher energy level down to n-1. Use the Rydberg equation to iden- tify the transition. c. to Dfinal ninitial POSTL

Explanation / Answer

E= h c / WL

h = Planck Constant = 6.626*10^-34 J s

c = speed of particle (i.e. light) = 3*10^8 m/s

E = energy per particle J/photon

WL = wavelength in meters

E= (6.626*10^-34)(3*10^8)/(121.58*10^-9) = 1.634*10^-18 J/photon

b)

Energy per mol

1.634*10^-18 J/photon * 6.022*10^23 photons/mol

E = 983994.8 J/mol

c)

1.634*10^-18 J/photon

Einitial = 2.178*10^-18

Apply Rydberg Formula

E = R*(1/nf^2 – 1/ni ^2)

R = -2.178*10^-18 J

Nf = final stage/level

Ni = initial stage/level

E = Energy per unit (i.e. J/photon)

E = R*(1/nf^2 – 1/ni ^2)

( 1.634*10^-18 ) = (2.178*10^-18) * (1/nf^2 - 1)

( 1.634*10^-18 ) / (2.178*10^-18)= (1/nf^2 - 1)

0.7502-1 = 1/nf^2

nf = (0.25^-1)^0.5

nf = 2

level 2

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