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Question

What are the frequency and wavelength of a photon emitted during a transition from n=5 state to the n=2 state in the hydrogen atom?

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Solution

Calculation for emitted energy

En=2.18×1018×z2n2J

Δ E=EfEi=2.18×10181n2f1n2f

Where, ni initial energy~(orbit=5)
nf final energy~(orbit=2)

(This transition gives rise to a spectral line in the visible region of the Balmer series.)

Δ E=2.18×1018[152122] J

=2.18×1018[42525×4]J

=2.18×1018×21100J

Δ E=4.58×1019J
It is the emission energy,now we have to find out correspondig frequency and wavelength

Calculation for frequency (v)

We know that ΔE=hv=hcλ

v=ΔEh

Where,h=planck constant=6.626×1034Js
ΔE=emission energy=4.58×1019J
v=?(Frequency)
Susbstituting the values in the equation, we get

v=4.58×1019J6.626×1024Js=0.691×1019+34

v=6.91×1014Hz(sec1)

Calculation for wavelength (λ)

We know, ΔE=hv=hcλ{v=cλ}
Where,c velocity of the light=3×108ms1

λ wavelength of the photon=?

λ =hcΔE=6.626×1034Js×3×108ms14.58×1019J

λ = 19.878×1034+84.58×1019m

λ = 4.34×1034+8+19=4.34×107m

λ = 434×109=434nm {1 nm=109m}

Final answer: v=6.91×1014Hz
λ = 434nm

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