What is the energy difference between the two quantum states involved in the transition of red light with wavelength 705 nm being absorbed by an atomic gas?

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What is the energy difference between the two quantum states involved in the transition of red light with wavelength 705 nm being absorbed by an atomic gas?

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Answer 1 · 2015-06-22T20:13:04

I found:
$E = h \cdot ν = 2.82 \times 10^{- 19} J = 1.76 e V$ $E=h*nu=2.82xx10^-19J=1.76eV$

Explanation:

A wavelength of $705 n m$ $705 nm$ corresponds to a frequency:
$ν = \frac{c}{λ} = \frac{3 \times 10^{8}}{7.05 \times 10^{- 7}} = 4.25 \times 10^{14} H z$ $nu=c/lambda=(3xx10^8)/(7.05xx10^-7)=4.25xx10^14Hz$

From Einstein's Relatonship you have that the photon energy will be:

$E = h \cdot ν = 6.63 \times 10^{- 34} \cdot 4.25 \times 10^{14} = 2.82 \times 10^{- 19} J = 1.76 e V$ $color(red)(E=h*nu)=6.63xx10^-34*4.25xx10^14=2.82xx10^-19J=1.76eV$

This is exactly the energy gap (or difference) between the two quantum states involved in the absorption of your photon of red light.

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What is the energy difference between the two quantum states involved in the transition of red light with wavelength 705 nm being absorbed by an atomic gas?

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