Question

A ground state hydrogen atom absorbs a photon of light having a wavelength of 93.73 nm. It then gives off a photon having a wavelength of 410.1 nm. What is the final state of the hydrogen atom? Values for physical constants can be found in the Chempendix.

Answer 1

Answer:

Final state of the hydrogen atom is the state where principle quantum number n = 2

Explanation:

When a photon of light having wavelength of 93·73 nm falls on hydrogen atom, the atom absorbs an energy of (h×c)÷93·73 nm

as Energy E = (h×c)÷wavelength

where

h is the Planck's constant

c is the speed of light in vaccum  which is 3×$10^{8}$ m/s

Energy that is released is (h×c)÷410·1 nm

Let the principle quantum  number of the final state of the hydrogen atom be n

According to the Bohr's principle

Energy difference between two states = ΔE = $R_{E}$ ( (1÷n²) - 1)

as the principle quantum number of the final state is n and principle quantum number of the initial state is 1

∴ ΔE = (h×c)((1÷93·73 nm) - (1÷410·1 nm))

$R_{E}$ ( (1÷n²) - 1) = (h×c)((1÷93·73 nm) - (1÷410·1 nm))

( $R_{E}$ ( (1÷n²) - 1)) ÷ (h×c) = (1÷93·73 nm) - (1÷410·1 nm)

∴ n≈2

∴ Principle quantum number of the final state is 2