Question

In the Bohr model, which of the following electron transitions in a hydrogen atom results in the emission of the highest-energy photon?
n = 3 to n = 2 OR n = 4 to n =3

<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

Answer 1

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.