## A monochromatic laser is exciting hydrogen atoms from the n=2 state to the n=5 state. part a what is the wavelength λ of the laser?

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## Answers ( 1 )

The energy levels of the hydrogen atom are given by

[tex]E_n=-13.6 frac{1}{n^2} [eV] [/tex]

where n is the level number.

In order to make transition from n=2 state to n=5 state, the electron should acquire an energy equal to the difference between the two energy levels:

[tex]Delta E= E_5 – E_2 = -13.6 frac{1}{5^2}-(-13.6 frac{1}{2^2})= -0.54+3.4=2.86 eV[/tex]

Keeping in mind that[tex]1 eV = 1.6 cdot 10^{-19}eV[/tex], we can convert this energy in Joules

[tex]Delta E = 2.86 eV cdot 1.6 cdot 10^{-19} J/eV=4.58 cdot 10^{-19} J[/tex]

This is the energy the photons of the laser should have in order to excite electrons from n=2 state to n=5 state. Their frequency can be found by using

[tex]Delta E=hf[/tex]

where h is the Planck constant and f is the photon frequency. Re-arranging it, we find

[tex]f= frac{Delta E}{h}= frac{4.58 cdot 10^{-19} J}{6.6 cdot 10^{-34} Js}=6.94 cdot 10^{14} Hz [/tex]

and finally we can use the relationship between frequency, wavelength and speed of light which holds for photons, in order to find their wavelength:

[tex]lambda= frac{c}{f}= frac{3 cdot 10^8 m/s}{6.94 cdot 10^{14} Hz}=4.32 cdot 10^{-7} m =432 nm [/tex]

and this is the laser wavelenghth.