Question

A ground state hydrogen atom absorbs a photon of light having a wavelength of 92.27 nm. It then gives off a photon having a wavelength of 383.4 nm. What is the final state of the hydrogen atom? Values for physical constants can be found here. nf=

please try to show solution

Answer #1

if in ground state.. assume n = 1

so..

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 = (-2.178*10^-18)*(1/nf^2 – 1/1 ^2)

clauclate Energy with the wavelength:

WL = h c / E

E1 = hc/WL1 = (6.636*!0^-34)(3*10^8)/(92.27*10^-9) =2.15758*10^-18

E2 = hc/WL2 = (6.636*!0^-34)(3*10^8)/(383.4*10^-9) =5.192488*10^-19

dE = E1-E2 = 2.15758*10^-18 - 5.192488*10^-19 = 1.6383*10^-18

then

E = (-2.178*10^-18)*(1/nf^2 – 1/1 ^2)

1.6383*10^-18 = (-2.178*10^-18)*(1/nf^2 – 1/1 ^2)

solve for nf

(1.6383*10^-18 ) / (-2.178*10^-18) = 1/nf^2 – 1

-0.75220 + 1 = 1/nf^2

nf = 1/0.25

nf = 2

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