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)

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
http://hyperphysics.phy-astr.gsu.edu/hbase/Bohr.html