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× 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