An electron in the n = 6 level of the hydrogen ato...

Question

An electron in the n = 6 level of the hydrogen atom relaxes to a lower energy level, emitting light of λ = 410 nm. {
m ;
m nmFind the principal level to which the electron relaxed.

Jules Bruno · Accepted Answer

We’re being asked to calculate for the principal level to which the electron relaxed.We’re going to use the Balmer Equation which relates wavelengths to a photon’s electronic transitions.1λ=RZ21n2final-1n2initialλ = wavelength, mR = Rydberg constant = 1.097x107 m-1 Z = atomic number of the elementninitial = initial energy levelnfinal = final energy levelCalculate the final energy level (nfinal):[readmore]For the electron to emit light and relaxed, it had to be excited. It must come from a higher energy level to a lower energy level.nfinal < 6Given:λ = 410 nm → convert m (1 nm = 10-9 m)Z = atomic number of Hydrogen = 1 (refer to the periodic table)ninitial = 6R = 1.097x107 m-1λ=410 nm×10-9 m1 nmλ = 4.10x10-7 m1λ=RZ21n2final-1n2initial14.10×10-7 m=1.097×107 m-1121n2final-1622.439×106 m-11.097×107 m-1=1.097×107 m-11n2final-1361.097×107 m-10.2223=1n2final-0.0277+0.0277            +0.0277n2final0.2500=1n2finaln2finaln2final0.25000.2500=10.2500n2final=4nfinal = 2.00The energy level to which the electron relaxed is at n = 2.

Problem: An electron in the n = 6 level of the hydrogen atom relaxes to a lower energy level, emitting light of λ = 410 nm. { m ; m nmFind the principal level to which the electron relaxed.

FREE Expert Solution

Problem Details

Problem: An electron in the n = 6 level of the hydrogen atom relaxes to a lower energy level, emitting light of λ = 410 nm. { m ; m nmFind the principal level to which the electron relaxed.

FREE Expert Solution

Problem Details

Sign up for free to watch this video!