Respuesta :
We can find the lower level by using Rydberg's formula: 1/w = R(1/L² - 1/U²) where:
w - wavelength (9.38nm),
L - lower energy level
U - upper energy level (6); and
R - Rydberg's constant (10,967,758 waves per meter for hydrogen).
So plugging in the information above:
1/ (9.38 * 10**-8 m.) = 10,967,758(1/L² - 1-36)
But we need to solve for L.
The left side (1/9.38 * 10**-8) becomes 10,660,980
So translating:
10660980 = 10967758(1/L² - 0.02777777)
Then divide both sides by 10967758
we get: 0.9720291 = 1/L² - 0.02777777.
then add 0.02777777 to both sides
we get: 0.9998 = 1/L²
0.9998 is very close to 1, so we can approximate 1/L² = 1 so L = 1.
Therefore the lower level is 1 which is the ground state.
w - wavelength (9.38nm),
L - lower energy level
U - upper energy level (6); and
R - Rydberg's constant (10,967,758 waves per meter for hydrogen).
So plugging in the information above:
1/ (9.38 * 10**-8 m.) = 10,967,758(1/L² - 1-36)
But we need to solve for L.
The left side (1/9.38 * 10**-8) becomes 10,660,980
So translating:
10660980 = 10967758(1/L² - 0.02777777)
Then divide both sides by 10967758
we get: 0.9720291 = 1/L² - 0.02777777.
then add 0.02777777 to both sides
we get: 0.9998 = 1/L²
0.9998 is very close to 1, so we can approximate 1/L² = 1 so L = 1.
Therefore the lower level is 1 which is the ground state.
The principal energy level to which the electron relaxed is [tex]\boxed{n=1}[/tex] .
Further Explanation:
When the transition of the electron from one energy level to another energy level takes place, there is an emission of the particular wavelength according to the energy of the level. The relation between the energy of emission and the wavelength is given as:
[tex]E=\frac{{hc}}{\lambda }[/tex]
Substitute the value of the Planck’s constant, speed of light and the wavelength of the emitted light in the above expression.
[tex]\begin{aligned}E&=\frac{{\left({6.63\times{{10}^{-34}}}\right)\times\left({3\times{{10}^8}}\right)}}{{93.8\times{{10}^{-9}}}}\\&=2.12\times{10^{-18}}\,{\text{J}}\\\end{aligned}[/tex]
This amount of energy is the change in the energy level as the electronic transition takes place.
The energy of the different energy levels in a hydrogen atom is expressed as:
[tex]\Delta E=13.6\,{\text{eV}}\left({\frac{1}{{n_f^2}}-\frac{1}{{n_i^2}}}\right)[/tex]
Here, [tex]\Delta E[/tex] is the change in the energy, [tex]{n_f}[/tex] is the final energy level and [tex]{n_i}[/tex] is the initial energy level.
Substitute the values in the above expression.
[tex]\begin{aligned}2.12\times{10^{-18}}&=13.6\times\left({1.6\times{{10}^{-19}}}\right)\left({\frac{1}{{n_f^2}}-\frac{1}{{{6^2}}}}\right)\\\frac{{2.12\times{{10}^{-18}}}}{{2.176\times{{10}^{-18}}}}&=\frac{1}{{n_f^2}}-\frac{1}{{36}}\\\frac{1}{{n_f^2}}&=0.97+\frac{1}{{36}}\\{n_f}&\approx 1\\\end{aligned}[/tex]
Thus, the principal energy level to which the electron relaxed is [tex]\boxed{n=1}[/tex] .
Learn More:
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Answer Details:
Grade: High School
Subject: Physics
Chapter: Hydrogen spectrum
Keywords:
Electron, level n=6, hydrogen atom, transition of electron, lower energy level, emitting light, wavelength, 93.8 nm, principal level, electron relaxed.
