Hydrogen exhibits several series of line spectra in different spectral regions. For example the Lyman series (nf = 1 in Balmer-Rydberg equation) occurs in the ultraviolet region while the Balmer (nf = 2) series occurs in the visible range and the Paschen (nf = 3), Brackett (nf = 4) and Pfund ( nf = 5) series all occur in the infrared range. What is the wavelength (in nm) of a line in the Brackett series where ni = 11?

E = 2.180E-18(1/4^2 - 1/11^2)

Solve for E, then
E = hc/wavelength and solve for wavelength (in meters), then convert to nm.

To find the wavelength (in nm) of a line in the Brackett series where ni = 11, we need to use the Balmer-Rydberg equation:

1/λ = R (1/ni^2 - 1/nf^2)

Given that nf = 4 (for the Brackett series) and ni = 11, we can substitute these values into the equation:

1/λ = R (1/11^2 - 1/4^2)

To calculate the wavelength, we need to know the value of the Rydberg constant (R) for hydrogen. The Rydberg constant for hydrogen is approximately 1.097 x 10^7 m⁻¹.

Converting the value of R to nm⁻¹:

R_nm = 1.097 x 10^7 m⁻¹ * 10⁹ nm/m

R_nm = 1.097 x 10^16 nm⁻¹

Now, we can substitute all the values into the equation and solve for 1/λ:

1/λ = (1.097 x 10^16 nm⁻¹) * (1/11^2 - 1/4^2)

1/λ = (1.097 x 10^16 nm⁻¹) * (1/121 - 1/16)

1/λ = (1.097 x 10^16 nm⁻¹) * (16/1936 - 121/1936)

1/λ = (1.097 x 10^16 nm⁻¹) * (-105/1936)

1/λ = -1.097 x 10^16 nm⁻¹ * 105/1936

1/λ = -1.097 x 10^16 nm⁻¹ * 105/1936

1/λ ≈ -6 x 10^12 nm⁻¹

To find λ, we will take the reciprocal of both sides:

λ ≈ 1/(-6 x 10^12 nm⁻¹)

λ ≈ -1.67 x 10^-13 nm

Since wavelength cannot be negative, disregard the negative sign and the final wavelength is approximately 1.67 x 10^-13 nm.

Therefore, the wavelength of a line in the Brackett series where ni = 11 is approximately 1.67 x 10^-13 nm.