Using Rydberg's equation for the energy levels (E = -R/n^2) and the wavelength and the c=(lambda)v, E=hv equations, SHOW/ PROVE how E(red band in H-spectrum, lambda = 656 nm) = (Delta or change/ difference) E (3 --> 2)?

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Using Rydberg's equation for the energy levels (E = -R/n^2) and the wavelength and the c=(lambda)v, E=hv equations, SHOW/ PROVE how E(red band in H-spectrum, lambda = 656 nm) = (Delta or change/ difference) E (3 --> 2)?

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Answer 1 · 2015-12-16T15:46:05

This is a bit long and probably there is a faster way but I tried this:

Explanation:

Ok; you know that when an electron jumps from an allowed orbit to another it absorbs/emits energy in form of a photon of energy $E=hnu$ (where $h=$ Planck's Constant and $nu=$ frequency).

Your "red" photon represents a transition between two orbits (of quantum numbers $n$ and $n+1$ ) separated by a "energy" of:

$E_(red)=h*nu_(red)$

but red means a wavelength: $lambda_(red)=656nm$

so the "red" frequency will be: $nu_(red)=c/lambda_(red)=(3xx10^8)/(656xx10^-9)=4.57xx10^14Hz$

And energy:

$E_(red)=6.63xx10^-34*4.57xx10^14=3xx10^-19J=1.87eV$
(where $1eV=1.6xx10^-19J$ )

Now we need to find the two orbits (their quantum numbers) from where and toward where the electron jumped:
We use Rydberg's Formula where we know that:
$DeltaE=E_f-E_i=1.87eV$
and also:
$DeltaE=-R(1/(n+1)^2-1/n^2)$
$1.87=-13.6(1/3^2-1/2^2)$
(I used Rydberg Constant in $eV$ ; $R=13.6eV$ )
I found:
$1.87=1.88$ that works fine I think!

Hope it helps!

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Using Rydberg's equation for the energy levels (E = -R/n^2) and the wavelength and the c=(lambda)v, E=hv equations, SHOW/ PROVE how E(red band in H-spectrum, lambda = 656 nm) = (Delta or change/ difference) E (3 --> 2)?

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Explanation:

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