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" />

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.

conduction is the transfer of thermal energy (heat in transfer) due to collisions between the molecules in the object. collisions between adjacent atoms and molecules transfer kinetic energy from the warmer to the cooler object. the objects must be in physical contact.