Find the wavelength of the radiation emitted when a hydrogen atom makes a transition from the n = 5 to the n =

Find the wavelength of the radiation emitted when a hydrogen atom makes a transition from the n = 5 to the n = 2 state.
(in nm)

The total energy of the nth orbit of a hydrogen atom is given by:

E_n = ½ k e² /r_n

with, radius of nth orbit: r_n = n² ro,

ro = 0.529 Angstron is the Bohr radius.

If an electron transitions from a higher orbit to a lower orbit it will emit a photon,
since it has to decrease its energy to match the binding energy of the new orbit.
Therefore, the energy carried away by the radiation is equal to the difference
between the two energy levels.

E = E_n1 - E_n2 = -(½ k e²/ro)[1/(n1)² - 1/(n2)²] = hc/ λ

1/λ =-(½ k e²/hcro)[1/(n1)² - 1/(n2)²]

where;
energy levels: n1 = 5, n2 = 2, Bohr radius: ro = 0.529x10^-10 m,
electron charge: e = 1.6x10^-19 C, Planck's constant: h = 6.6x10^-34 J-s,
electrostatic constant: k = 9x10^9 N m²/C², speed of light: c = 3x10^8 m/s

1/λ = {[(0.5*9x10^9)(1.6²x10^-38)] / [(6.6x10^-34)(3x10^8)(0.529x10^-10)]}(1/… - 1/5²)

1/λ = 2.3x10^8 => λ = 4.3 nm