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What is the energy of a photon that, when absorbed by a hydrogen atom, could cause an electronic transition from the n=2 state to the n=5 state?

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To determine the energy of a photon required for an electronic transition from the n=2 state to the n=5 state in a hydrogen atom, we can use the formula for the energy of a photon:

E = ΔE = hc/λ

Where:

- E is the energy of the photon

- ΔE is the change in energy between the initial and final states

- h is Planck's constant (approximately 6.626 x 10^-34 joule-seconds)

- c is the speed of light (approximately 3 x 10^8 meters per second)

- λ is the wavelength of the photon

The energy difference between two energy levels in a hydrogen atom is given by the Rydberg formula:

ΔE = Rh * (1/n_f^2 - 1/n_i^2)

Where:

- ΔE is the change in energy

- Rh is the Rydberg constant (approximately 2.18 x 10^-18 joules)

- n_f is the final energy level (n=5 in this case)

- n_i is the initial energy level (n=2 in this case)

Substituting the values into the Rydberg formula:

ΔE = Rh * (1/5^2 - 1/2^2)

= Rh * (1/25 - 1/4)

= Rh * (4/100 - 25/100)

= Rh * (-21/100)

≈ -0.0218 * Rh

Now, we can substitute this change in energy value into the energy formula for the photon:

E = hc/λ = -0.0218 * Rh

Rearranging the equation to solve for λ:

λ = hc / E

Substituting the values for h, c, and E:

λ = (6.626 x 10^-34 joule-seconds * 3 x 10^8 meters per second) / (-0.0218 * Rh)

Calculating this expression will give us the wavelength of the photon required for the electronic transition.

User Lukas Gentele
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