Calculate the wavelength (in nm) of a photon emitted during a transition corresponding to the first line in the Balmer series (nf = 2) of the hydrogen emission spectrum.
"Balmer Series": from n = 5 to n = 2
energy of the photon with the frequency ν
E=hν
h = Planck's constant
For any state
En=−13.6 eV/n2
where −13.6 eV = ground-state energy of the hydrogen atom.
E2=−13.6 eV/4
=−3.4 eV
=−5.44 x 10^−19 J
E5 =−13.6 eV/25
=0.544 eV
=−8.7x 10^−20 J
So,
ΔE= E5 - E2
= 4.57 x 10^−19 J.
From the equation
E=hν
ν = (4.57 x 10^−19 J) /( 6.63 x10^−34J⋅s)
= 6.892 x 10^14 s−1
Speed of light c=λν
So,
λ= (3 x 10^8 m/s) / (6.892 x 10^14 s−1) × (10^9 nm/1 m)
=435 nm
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