Question

What is the energy of light associated with a transition from n=3 to n=8 in a hydrogen atom? Does this represent absorption or emission of a photon?

Answer 1

Explanation:

Energy levels to be n = 8 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition

1λ = R ⋅ (1/nf^2 − 1/ni^2)

where,

λ - the wavelength of the emitted photon

R - Rydberg's constant = 1.0974 x 10^7m

nf - the final energy = 8

ni - the initial energy level = 3

1/λ = 1.0974 x 10^7 * (1/8^2 − 1/3^2)

= -1.05x 10^6 m.

Using Heinsberg's equation,

E = (h * c)/λ

Calculating the energy of this transition you'll have to multiply Rydberg's equation by h * c

where,

h - Planck's constant = 6.626 x 10^−34 Js

c - the speed of light = 3.0 x 10^8 m/s

So, the transition energy, E = (6.626 x 10^−34 * 3 x 10^8) * -1.05x 10^6

= -2.08 x 10^-19 J.

B.

When an electron transitions from a less excited state to a excited state (higher energy orbit), the difference in energy is absorbed as a photon.

The energy is negative which means energy is lost or dissipated to the surroundings. Therefore, an absorption of photons.