a ground state hydrogen atom absorbs a photon of light Having a wavelength of 94

Question

a ground state hydrogen atom absorbs a photon of light Having a wavelength of 94.92 nm. It then gives off a photon having a wavelength of 4.050x10^3 nm. What is the final state of the hydrogen atom? a ground state hydrogen atom absorbs a photon of light Having a wavelength of 94.92 nm. It then gives off a photon having a wavelength of 4.050x10^3 nm. What is the final state of the hydrogen atom? a ground state hydrogen atom absorbs a photon of light Having a wavelength of 94.92 nm. It then gives off a photon having a wavelength of 4.050x10^3 nm. What is the final state of the hydrogen atom?

Explanation / Answer

we know that

energy = hc / lamda

so

energy in ground state (Eo) = 6.634 x 10-34 x 3 x 10^8 / 94.92 x 10-9

Eo = 2.0967 x 10-18 J

also

1 ev = 1.6 x 10-19 J

so

Eo = 2.0967 x 10-18 / 1.6 x 10-19   eV

Eo = 13.104 eV

now

in the final state

Ef = 6.634 x 10-34 x 3 x 10^8 / 4.05 x 1000 x 10^-9

Ef = 4.914 x 10-20 J

Ef = 4.914 x 10-20 / 1.6 x 10-19

Ef = 0.307 eV

now

the energey difference

E = 13.104 -0.307

E = 12.797

we know that

for

hydrogen atoms

Ef = -13.6 / (nf)^2 eV

so

we get

-13.6/(nf)^2 - ( -13.6 / 1) = 12.797

nf = 4

so

the final state of the hydrogen atom is 4

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