Question

A hydrogen atom in the n=7 state decays to the n=4 state. what is the wavelength of the photon that the hydrogen atom emits? use hc=1240 nm ev

Answer 1

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, brainly.com/question/2393994

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