A hydrogen atom transitions from the n=3 to the n=2 states by emitting a photon. What is the wavelength of the photon? (Estimate without a calculator)

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The Bohr formula gives `E=E_0/n^2` for the nth excited state with `E_0=13.6 eV` being the ground state energy.

The relationship between energy and wavelength is

`Delta E=hc/(Delta lambda)`

`Delta lambda=(hc)/(Delta E)=(hc)/(E_0/n_f^2-E_0/n_i^2)=(hc)/E_0*(1/n_f^2-1/n_i^2)^-1`

A nice thing to know is `barh*c~~200 MeV*fm=2*10^-7 eV*m` , where `barh` is the reduced planks constant.

`Delta lambda...

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The Bohr formula gives `E=E_0/n^2` for the nth excited state with `E_0=13.6 eV` being the ground state energy.

The relationship between energy and wavelength is

`Delta E=hc/(Delta lambda)`

`Delta lambda=(hc)/(Delta E)=(hc)/(E_0/n_f^2-E_0/n_i^2)=(hc)/E_0*(1/n_f^2-1/n_i^2)^-1`

A nice thing to know is `barh*c~~200 MeV*fm=2*10^-7 eV*m` , where `barh` is the reduced planks constant.

`Delta lambda ~~(2pi*2*10^-7 eV*m)/(13.6 eV)*(1/4-1/9)^-1`

`Delta lambda =(2pi*2*10^-7 eV*m)/(13.6 eV)*36/5`

`Delta lambda ~~ (6*2*10^-7 eV*m)/(14 eV)*7`

`Delta lambda ~~(12*10^-7 eV*m)/(2 eV)`

`Delta lambda ~~6*10^-7 m`

`Delta lambda =600*nm`

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