Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
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
#SPJ4
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
