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A hydrogen atom in the n=7 state decays to the n=4 state. The wavelength of the photon that the hydrogen atom emits is 4592.59nm.
The Energy of photon is the energy possessed by a photon when it moves from a high energy level to a low energy level. It emits a photon of a certain wavelength. The following relation can be used to find out the relation between the energy levels and the energy possessed:
E = 13.6 × Z² (1/n₂² - 1/n₁²) eV
where, n₁ is the initial energy level i.e. n₁ =7
n₂ is the higher energy level i.e. n₂ = 4
E is the energy possessed
Z is the atomic number, Z = 1 for H-atom
Subsituting in above equation,
E = 13.6 (1/16 - 1/49) eV
E = 0.27 eV
We know that,
E = hc / λ
where, h is Planck constant
c is speed of light
λ is wavelength
On subsituting,
0.27 eV = 1240/ λ
⇒ λ = 4592.59 nm
Hence, the wavelength of photon emitted by Hydrogen atom is 4592.59nm.
Learn more about Energy of Photon here, https://brainly.com/question/2393994
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