a) According to the Rydberg equation, the line with the shortest wavelength in the emission spectrum of atomic hydrogen is predicted to lie at a wavelength (in nm) of _____
b) According to the Rydberg equation, the longest wavelength (in nm) in the series of H-atom lines produced when the electron falls to the state n = 3 is ____
1/wavelength = R(1/n^2 - 1/n^2) where the two n symbols are n1 for the first one and n2 for the second.
So the shortest wavelength (the most energy) will come when an electron falls from the outside most energy level to the inner most energy level. Therefore, plug in n1 = 1 and n2 = 0 and solve for wavelength. That will be the Lyman series limit which you probably have in your text.
2. The longest wavelength when n1 = 3 (it falls to 3) would be for it to start at n =4. So n1 = 3 and n2 = 4. Remember n1<n2 so in my mind this is backwards but you gotta go with the flow. It's because the energy levels have - energy and when the electron is at an infinite distance that's when energy = 0
To find the answer to both questions, we need to use the Rydberg equation. The Rydberg equation is given as:
1/λ = R_H (1/n₁² - 1/n₂²)
Where:
λ is the wavelength of the emitted light,
R_H is the Rydberg constant for hydrogen (approximately 1.097 × 10^7 m⁻¹),
n₁ is the initial energy level (or principal quantum number),
n₂ is the final energy level (or principal quantum number).
a) For the line with the shortest wavelength in the emission spectrum of atomic hydrogen, we need to find the combination of n₁ and n₂ that produces the smallest wavelength.
In this case, we start with the electron in the ground state (n = 1) and calculate the wavelength when it falls to the next lowest energy level (n = 2). Plugging these values into the Rydberg equation:
1/λ = R_H (1/1² - 1/2²)
Simplifying:
1/λ = R_H (1 - 1/4)
1/λ = R_H * (3/4)
λ = 4R_H/3
To convert this wavelength into nm (nanometers), multiply the value by 10^9 to account for the conversion factor between meters and nanometers:
λ = 4R_H/3 * 10^9 nm
b) For the longest wavelength in the series of hydrogen atom lines produced when the electron falls to the state n = 3, we follow a similar approach.
To find the wavelength, we plug in n₁ = 3 and n₂ = 1 into the Rydberg equation:
1/λ = R_H (1/3² - 1/1²)
Simplifying:
1/λ = R_H (1/9 - 1)
1/λ = R_H * (1 - 9/9)
1/λ = R_H/9
λ = 9R_H
Again, we convert this wavelength into nm:
λ = 9R_H * 10^9 nm
Calculating the specific values for R_H and performing the multiplication will provide the answers to both questions.
a) The Rydberg equation is given as:
1/λ = R * (1/n1^2 - 1/n2^2)
where λ is the wavelength, R is the Rydberg constant, n1 is the initial energy level, and n2 is the final energy level.
For the line with the shortest wavelength in the emission spectrum of atomic hydrogen, the final energy level is n2 = 1.
Plugging this value into the equation, we have:
1/λ = R * (1/n1^2 - 1/1^2)
1/λ = R * (1/n1^2 - 1)
1/λ = R/n1^2 - R
We know that the Rydberg constant, R, is 1.097373 x 10^7 m⁻¹. To convert this to nm⁻¹, we multiply by 10^9 (since there are 10^9 nm in 1 m):
R = 1.097373 x 10^7 x 10^9 nm⁻¹
R = 1.097373 x 10^16 nm⁻¹
Now, let's consider the limit where the wavelength is the shortest. That means n1 is very large. Taking the limit as n1 approaches infinity, we get:
1/λ = 0 - R
1/λ = -R
λ = -1/R
Now, plug in the value for R:
λ = -1/(1.097373 x 10^16 nm⁻¹)
λ ≈ -9.1093838 x 10^-17 nm⁻¹
Since the wavelength cannot be negative, we disregard the negative sign. Therefore, the line with the shortest wavelength in the emission spectrum of atomic hydrogen is approximately 9.1093838 x 10^-17 nm⁻¹.
b) According to the Rydberg equation, the longest wavelength in the H-atom series occurs when the electron falls to the state n = 3.
Using the Rydberg equation again:
1/λ = R * (1/n1^2 - 1/n2^2)
Given that n2 = 3, we can calculate the wavelength by setting n1 to different values and finding the corresponding wavelengths for the longest wavelength in the series.
Let's plug n1 = 4 into the equation:
1/λ = R * (1/4^2 - 1/3^2)
Simplifying:
1/λ = R * (1/16 - 1/9)
1/λ = R * (9/144 - 16/144)
1/λ = R * (-7/144)
Now, plug in the value for R (1.097373 x 10^16 nm⁻¹):
1/λ = -7 * 1.097373 x 10^16 / 144 nm⁻¹
1/λ ≈ -5.3529939 x 10^14 / nm
Again, we ignore the negative sign:
λ = 1/ (5.3529939 x 10^14) nm
λ ≈ 1.868 x 10^-15 nm
Therefore, the longest wavelength in the series of H-atom lines produced when the electron falls to the state n = 3 is approximately 1.868 x 10^-15 nm.