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The n = 2 to n = 6 transition in the bohr hydrogen atom corresponds to the ________ of a photon with a wavelength of ________ nm.

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skyluke89
The energy levels of the hydrogen atom are quantized and their energy is given by the approximated formula
[tex]E=- \frac{13.6}{n^2} [eV] [/tex]
where n is the number of the level.

In the transition from n=2 to n=6, the variation of energy is
[tex]\Delta E=E(n=6)-E(n=2)=-13.6 ( \frac{1}{6^2}- \frac{1}{2^2} )[eV]=3.02 eV[/tex]
Since this variation is positive, it means that the system has gained energy, so it must have absorbed a photon.

The energy of photon absorbed is equal to this [tex]\Delta E[/tex]. Converting it into Joule,
[tex]\Delta E=3.02 eV=4.84 \cdot 10^{-19}J[/tex]
The energy of the photon is
[tex]E=hf[/tex]
where h is the Planck constant while f is its frequency. Writing [tex]\Delta E=hf[/tex], we can write the frequency f of the photon:
[tex]f= \frac{\Delta E}{h}= \frac{4.84 \cdot 10^{-19}J}{6.63 \cdot 10^{-34}m^2 kg/s}=7.29 \cdot 10^{14}Hz [/tex]

The photon travels at the speed of light, [tex]c=3 \cdot 10^8 m/s[/tex], so its wavelength is
[tex]\lambda = \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{7.29 \cdot 10^{14}Hz}=4.11 \cdot 10^{-7}m=411 nm [/tex]

So, the initial sentence can be completed as:
The n = 2 to n = 6 transition in the bohr hydrogen atom corresponds to the "absorption" of a photon with a wavelength of "411" nm.
pstnonsonjoku

The  n = 2 to n = 6 transition in the Bohr hydrogen atom corresponds to the absorption of a photon with a wavelength of 410 nm.

Given the Rydberg formula;

1/λ = R(1/n^2final - 1/n^2initial)

λ = wavelength of the photon

R = Rydberg constant

n2 = final state

n1 = initial state

Substituting values;

1/λ = 1.097 × 10^ 7 m-1(1/3^2 - 1/6^2)

1/λ = 1.097 × 10^ 7 m-1 (0.25 - 0.027)

λ = 4.10 × 10-7 m or 410 nm

Learn more: https://brainly.com/question/15178305

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