Answer to Question #239903 in General Chemistry for helen

Question

Determine the wavelength of the light absorbed when an electron in a hydrogen atom makes a transition from an orbital in the 

Answer 1

Assuming the electron made a transition from an orbital n = 2 to an orbital in which n =5

The energy levels of the hydrogen atom are quantized and their energy is given by the approximated formula

$E=- \dfrac{13.6}{n^2}[eV]$

where n is the number of the level.

In the transition from n=2 to n=5, the variation of energy is

$\Delta E=E(n=5)-E(n=2)=-13.6 ( \dfrac{1}{5^2}- \dfrac{1}{2^2} )[eV]=2.86 eV$

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,

$\Delta E=2.86 eV\cdot (1.6 \cdot 10^{-19})=4.57 \cdot 10^{-19} J$

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:

$f= \dfrac{\Delta E}{h}= \dfrac{4.57 \cdot 10^{-19} J}{6.62 \cdot 10^{-34}Js}=6.9 \cdot 10^{14}Hz$

The photon travels at the speed of light,$c=3 \cdot 10^8 m/s$, so its wavelength is $\lambda = \dfrac{c}{f}= \dfrac{3 \cdot 10^8 m/s}{6.9 \cdot 10^{14}Hz} =4.35 \cdot 10^{-7}m = 435 nm$