Aground 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.

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× 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 = ( (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))

( (1÷n²) - 1) = (h×c)((1÷93·73 nm) - (1÷410·1 nm))

( ( (1÷n²) - 1)) ÷ (h×c) = (1÷93·73 nm) - (1÷410·1 nm)

∴ n≈2

∴ Principle quantum number of the final state is 2

the increase of protons   hope this