Answer:
n = 3 to n = 2
Explanation:
The Rydberg formula for electron transitions in the hydrogen atom is given by:
1/λ = Rh( 1/n₁² - 1/n₂² )
where
1/λ = wavelength of the transition,
Rh = Rydberg constant,
n₁ and n₂ are the principal quantum numbers of the energy levels involved in the transition with n₂ greater n₁.
The energy of the photon is given by
E = hc/λ
where
h= Planck´s constant
c= speed of light
Therefore the energy is inversely proportional to the wavelength and the term ( 1/n₁² - 1/n₂² ) will be greater ( 1/2² - 1/3² ) than (1/3² - 1/4² ) ( 0.14 vs 0.05 ). So the transition from n= 3 to n=2 will result in the emission of the highest energy photon in this question.
