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1. An electron transitions from the n = 6 to the n = 4 quantum state...

1. An electron transitions from the n = 6 to the n = 4 quantum state of the hydrogen atom. Is photon absorbed or emitted for the associated electron transition? What is the wavelength of the associated photon? Energy levels: En = -2.1810-18J ; Speed of light: c=3.00 ; Plank constant: h=6.63

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Answer #1

We know that,

En = -2.18*10^-18 / n^2

So, E6 = -2.18*10^-18 / 36 = -0.06055*10^-18 J

E4 = -2.18*10^-8 / 16 = -0.13625*10^-18 J

Delta E = E4 - E6 = -0.13625*10^-18 - (-0.06055*10^-18)

Delta E = -0.0757*10^-18 J

-ve sign shows energy is released i.e. photon is emitted.

Also,

Delta E = hc/L

L = hc/Delta E = 6.6*10^-34 * 3*10^8 / 0.0757*10^-18

L = 261.559*10^-8 meter

Wavelength = L = 2.615*10^-6 meters ..... Answer

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