An electron in an atom initially has an energy 5.5 eV above the ground state energy.It drops to a state with energy 3.2 eV above the ground state energy and emits a photon in the process.The wave associated with the photon has a wavelength of: A)5.4 *10 ^- ⁷ m B)3.0 *10 ^- ⁷ m C)1.7 *10 ^- ⁷m D)1.2*10 ^- ⁷ m E)1.0 *10 ^- ⁷m

Question

Quizplus · Accepted Answer

The energy difference between the initial state and the final state is:

ΔE = 5.5 eV - 3.2 eV = 2.3 eV

The energy of the photon emitted is equal to the energy difference, so:

E = ΔE = 2.3 eV

We can convert this to joules using the conversion factor 1 eV = 1.6 x 10^-19 J:

E = 2.3 eV x 1.6 x 10^-19 J/eV = 3.68 x 10^-19 J

The energy of a photon is related to its wavelength by the equation:

E = hc/λ

where h is Planck's constant and c is the speed of light. Rearranging this equation to solve for λ, we get:

λ = hc/E

Plugging in the values for h, c, and E that we have, we get:

λ = (6.63 x 10^-34 J s) x (3.00 x 10^8 m/s) / (3.68 x 10^-19 J) = 5.4 x 10^-7 m

Converting meters to nanometers, we get:

λ = 540 nm

Therefore, the wavelength of the photon emitted is 5.4 x 10^-7 m or 540 nm. Choice A is the correct answer.

An Electron in an Atom Initially Has an Energy 5