Energetic Electrons
In atoms, electrons jump between energy levels, and they absorb or radiate energy in the form of light (photons). An electron falls to a lower energy level when it radiates a photon, and jumps to a higher energy level when it absorbs a photon.
Find the wavelength of the photon that radiates when an electron jumps from the n = 3 energy level down to the n = 2 energy level. Where is this photon in the electromagnetic spectrum?
Assumptions
- h (Planck's constant) = 4 . 1 4 × 1 0 − 1 5 eV s.
- c (Speed of light) = 3 × 1 0 8 m/s.
- All values are approximate.
- Assume that this is a hydrogen atom.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
6 solutions
Discussions for this problem are now closed
You can also solve this problem using the Rydberg formula with Rydberg constant ( R ) = 1 . 0 9 7 × 1 0 7 m − 1 and the transitioning levels are from n 2 = 3 to n 1 = 2 and since here hydrogen atom is considered, so Z = Atomic no. of the considered element = 1 .
The Rydberg formula is like this ----->
λ v a c 1 = R Z 2 ( n 1 2 1 − n 2 2 1 )
This formula is only applicable for hydrogen-like chemical elements.
yes....thats true.prasun....how do find time for latex
It doesn't take time for me as I am habituated with writing in LaTeX.... ;)
Calculation is good, but with this question I happen to have H-alpha solar telescope. This works on photons with a wavelength 656,28 nm wich is emitted by an electron when it drops from 3. to 2. orbital.
You can use de Broglie wavlength infact.
E = − 1 3 . 6 / n 2 eV
h c / λ = − 1 3 . 6 ( 1 / n 1 2 − 1 / n 2 2 ) e V = − 1 3 . 6 ( 1 / 4 − 1 / 9 ) e V = − 1 3 . 6 × 5 / 3 6 e V
λ = h c / 1 3 . 6 × 3 6 / 5 = 6 5 6 nm
why is energy equal -13.6/n^2 eV i dont under stand please explain it for me
Total energy of an electron is the sum of kinetic and potential energy, which comes out to be = -13.6 (Z^2)/(n^2) eV (Z being the atomic number and n the shell) . Since it is hydrogen, Z=1.
from n-3 to n-2, -1,51-(-3,40) = 1,89 eV. E = (h.c)/λ 1,89eV.λ = h.c 1,89eV.λ = 4.14x10^-15eVs x 3x10^8m/s λ = 1.242^-6eV m/1,89eV λ = 6.57^-7m = 657nm 657nm is located on visible light range, and it is red
use hc=12400(app)
wavelenth of emitted photon can be determined using the formula = 12400/(E2-E1), where E2, E1 are energy levels. When an electron jumps from outer to 2nd orbit the it lies in visible region. Thus 12400/(-1.5+3.4)=6526*10^-10 =>652nm
Well λ = E h c . We just need to compare Planck's equation and v = λ ⋅ f
from level 3 to level 2 it's balmer series..so visible range. use equations E=hf and then c=lambda.f
From level 3 to level 2 there is a loss of 1.89eV as radiation. We can use E=hf to find the frequency and then c=λf to find the wavelength.