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

Answer 2

The final state of the hydrogen atom will be the first excited state, where n = 2.

What is the final state of the hydrogen atom?

The energy of the n-th state of the hydrogen atom is:

$E_n = \frac{-13.6 eV}{n^2}$

Particularly, for the ground state we have n = 1, so its energy is:

$E_1 = -13.6 eV}$

At this value, we must add the energy of a photon with a wavelength of 93.74 nm, and subtract the energy of a photon with a wavelength of 410.1 nm.

The energy of a photon is given by:

$E = \frac{h*c}{\lambda}$

Where:

• h = 6.63*10^{-34} J*s
• c = 3*10^8 m/s
• λ = wavelength.

Then the energies of the photons are:

$E_{93,73nm} = \frac{6.63*10^{-34} J*s*3*10^8 m/s}{93.73*10^{-9} m} = 2.12*10^{-18} J\\\\E_{93,73nm} = \frac{6.63*10^{-34} J*s*3*10^8 m/s}{410.1*10^{-9} m} = 4.85*10^{-19} J$

Where we used that:

1nm = 1*10^{-9} m

Then we use the conversion:

1 J = 6.24*10^18 eV to write the energies as:

$E_{93,73nm} = 2.12*10^{-18} J =2.12*10^{-18}*(6.24*10^{18} eV) = 13.23 eV \\\\E_{93,73nm} = 4.85*10^{-19} J = 4.85*10^{-19}*(6.24*10^{18} eV) = 3.03 eV$

Then the final energy of the hydrogen atom is:

$E = -13.6eV + 13.23 eV - 3.03 eV = -3.4 eV$

To get the final state, we must find n such that:

$E_n = \frac{-13.6 eV}{n^2} = -3.4 eV\\\\ n = \sqrt{13.6/3.4} = 2$

So the hydrogen atom is on the first excited state.

If you want to learn more about the hydrogen atom, you can read:

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