A radioactive sample has a count rate of 800 counts per minute. One hour later, the count rate has fallen to 100 counts per minu
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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]](https://tex.z-dn.net/?f=E%3D-%20%20%5Cfrac%7B13.6%7D%7Bn%5E2%7D%20%5BeV%5D%20)
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](https://tex.z-dn.net/?f=%5CDelta%20E%3DE%28n%3D6%29-E%28n%3D2%29%3D-13.6%20%28%20%5Cfrac%7B1%7D%7B6%5E2%7D-%20%5Cfrac%7B1%7D%7B2%5E2%7D%20%20%29%5BeV%5D%3D3.02%20eV)
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
. Converting it into Joule,
The energy of the photon is
where h is the Planck constant while f is its frequency. Writing
, we can write the frequency f of the photon:
The photon travels at the speed of light,
, so its wavelength is
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.
where n is the number of the level.
In the transition from n=2 to n=6, the variation of energy is
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
The energy of the photon is
where h is the Planck constant while f is its frequency. Writing
The photon travels at the speed of light,
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.