Answer:
A) 1875 nm (if we are talking about the series of H- atom lines with
n1 = 3 )
Explanation:
First we must bear in mind that being n1 = 3 we will be in the Paschen series with emission up to level n = 3 that could come from n = 4, n = 5.
In chemistry, the Paschen Series (also called the Ritz-Paschen Series) is the series of transitions and emission lines resulting from the hydrogen atom when electron jumps from a state of n ≥ 4 an = 3, where n refers to the quantum number main of the electron. Transitions are called sequentially with Greek letters: n = 4 to n = 3 is called Paschen-alpha, 5 to 3 is Paschen-beta, 6 to 3 is Paschen-gamma.
To obtain the longest wavelength for n1 = 3 it must be from the upper level closest to this level, therefore from the level n = 4. An electro that falls from a n = 5 an = 3 would release more energy and be of shorter wavelength .
We can perform the calculations using the Rydberg equation:
1/λ = Rh * [(1 / (n1)^2) - (1 / (n2)^2)]
So:
1/wavelength =
R [(1 /3^2) - (1 / 4^2)] =
R (1/9 - 1/16) =
R x 0.04861 = 533251.7 m
wavelength = 1/533251.7
wavelength = 1875m
Answer:
A) 1875 nm
Explanation:
Hello,
In this case, the Rydberg equation is shown below:
[tex]\frac{1}{\lambda} =R(\frac{1}{n_1^2}-\frac{1}{n_2^2})[/tex]
For the series of H- atom, one assigns: [tex]n_1=3[/tex] as this is the Paschen series with emission down to the [tex]n=3[/tex] level. Now, we must consider that the longest wavelength is from the nearest upper level to the [tex]n=3[/tex] level, that is, the [tex]n=4[/tex] level. In such a way, the resulting wavelength is computed as shown below:
[tex]\frac{1}{\lambda} =(10973731.5m^{-1}*\frac{1x10^{-9}m}{1nm} )(\frac{1}{3^2}-\frac{1}{4^2})\\\frac{1}{\lambda} =5.334x10^{-4}nm^{-1}\\\lambda=\frac{1}{5.334x10^{-4}nm^{-1}} \\\lambda=1875nm[/tex]
Best regards.
