Question

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.

Answer #1

"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/n^{2}

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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