Question

According to the Rydberg equation, the longest wavelength (in nm) in the series of H- atom lines with A) 1875 nm B) 1458 n Ans: A C) 820 nm D) 656 nm E) 365 nm

Answer 1

A) 1875 nm

Step-by-step explanation:

Hello,

In this case, the Rydberg equation is shown below:

`(1)/(\lambda) =R((1)/(n_1^2)-(1)/(n_2^2))`

For the series of H- atom, one assigns:
`n_1=3` as this is the Paschen series with emission down to the
`n=3` level. Now, we must consider that the longest wavelength is from the nearest upper level to the
`n=3` level, that is, the
`n=4` level. In such a way, the resulting wavelength is computed as shown below:

`(1)/(\lambda) =(10973731.5m^(-1)*(1x10^(-9)m)/(1nm) )((1)/(3^2)-(1)/(4^2))\\(1)/(\lambda) =5.334x10^(-4)nm^(-1)\\\lambda=(1)/(5.334x10^(-4)nm^(-1)) \\\lambda=1875nm`

Best regards.

Answer 2

A) 1875 nm (if we are talking about the series of H- atom lines with

n1 = 3 )

Step-by-step 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