The Hydrogen Atom: Photon Energy of n=4 to n=3 Transition

[UrbanXrisis]

What is the energy of the photon when the Hydrogen aton transitions from the n=4 to n=3 energy state?

 

[swansont]

What is the energy of the photon when the Hydrogen aton transitions from the n=4 to n=3 energy state?

It has E = E₄ - E₃ because energy is conserved.

Use the Bohr energy level equation to find the indiviual level values

 

[M98Ranger]

The energy of a photon emitted during the transition of a hydrogen atom from the n=4 to n=3 energy state can be calculated using the Rydberg formula:

E = -Rhc(1/n1^2 - 1/n2^2)

Where E is the energy of the photon, R is the Rydberg constant (1.097x10^-2 nm^-1), h is Planck's constant (6.626x10^-34 J*s), and c is the speed of light (2.998x10^8 m/s).

Plugging in the values for n1=4 and n2=3, we get:

E = -((1.097x10^-2 nm^-1)(6.626x10^-34 J*s)(2.998x10^8 m/s))(1/4^2 - 1/3^2)

= -2.178x10^-18 J

Therefore, the energy of the photon emitted during the n=4 to n=3 transition is 2.178x10^-18 J. This corresponds to a wavelength of approximately 911.7 nm, which falls within the infrared region of the electromagnetic spectrum.