Question

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

Answer 1

Answer: Absorption, 410

Explanation:

Answer 2

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

In the transition from n=2 to n=6, the variation of energy is
$\Delta E=E(n=6)-E(n=2)=-13.6 ( \frac{1}{6^2}- \frac{1}{2^2} )[eV]=3.02 eV$
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 $\Delta E$. Converting it into Joule,
$\Delta E=3.02 eV=4.84 \cdot 10^{-19}J$
The energy of the photon is
$E=hf$
where h is the Planck constant while f is its frequency. Writing $\Delta E=hf$, we can write the frequency f of the photon:
$f= \frac{\Delta E}{h}= \frac{4.84 \cdot 10^{-19}J}{6.63 \cdot 10^{-34}m^2 kg/s}=7.29 \cdot 10^{14}Hz$

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

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.