Calculate the wavelength in nm of the line with n = 6 in the Pfund series of the spectrum of atomic hydrogen.
Use the following formula:
Wavenumber = RH*((1/n1^2) - (1/n2)^2)
Where wavenumber = (1/wavelength),
RH = Rydberg's constant,
n1 = The number corresponding to the series (Pfund = 5)
n2 = The 'n' value of the line (n = 6)
To calculate the wavelength in nm of the line with n = 6 in the Pfund series of the spectrum of atomic hydrogen, we can use the following formula:
1/λ = R * (1/n1^2 - 1/n2^2)
where λ is the wavelength, R is the Rydberg constant (1.097373×10^7 m^-1), n1 is the lower energy level, and n2 is the higher energy level.
In this case, n1 = 5 and n2 = 6.
Plugging these values into the formula, we get:
1/λ = 1.097373×10^7 m^-1 * (1/5^2 - 1/6^2)
Simplifying the equation:
1/λ = 1.097373×10^7 m^-1 * (1/25 - 1/36)
1/λ = 1.097373×10^7 m^-1 * (11/900)
1/λ = 1.219303×10^4 m^-1
Now, we can solve for λ by taking the reciprocal of both sides:
λ = 1/1.219303×10^4 m^-1
λ ≈ 8.205 nm
Therefore, the wavelength of the line with n = 6 in the Pfund series of the spectrum of atomic hydrogen is approximately 8.205 nm.
To calculate the wavelength in nm of the line with n = 6 in the Pfund series of the spectrum of atomic hydrogen, we need to use the formula for calculating the wavelength of a spectral line:
1/λ = R_H * (1/n_final^2 - 1/n_initial^2),
where λ is the wavelength, R_H is the Rydberg constant for hydrogen, n_final is the final principal quantum number, and n_initial is the initial principal quantum number.
In this case, we have n_initial = 6 and n_final = ∞ (since the Pfund series corresponds to transitions where the electron goes from a higher energy level to the nf = 6 level). The Rydberg constant R_H for hydrogen is approximately 1.097 × 10^7 m^-1.
Plugging in the values into the formula, we get:
1/λ = (1.097 × 10^7 m^-1) * (1/6^2 - 1/∞^2)
Now, we need to convert the wavelength from meters (m) to nanometers (nm). By multiplying both sides of the equation by λ and rearranging the equation, we get:
λ = 1 / ((1.097 × 10^7 m^-1) * (1/6^2 - 1/∞^2))
To calculate λ, we need to consider the limit as n_final approaches infinity. When n_final is very large, 1/n_final^2 approaches zero and 1/∞^2 becomes negligible. Thus, we can ignore the second term in the equation, resulting in:
λ = 1 / (1.097 × 10^7 m^-1 * 1/6^2)
Calculating this expression, we get:
λ ≈ 1 / (1.097 × 10^7 m^-1 * 1/36)
λ ≈ 36 * 1.097 × 10^7 nm
λ ≈ 39.492 × 10^7 nm
Therefore, the wavelength of the line with n = 6 in the Pfund series of the spectrum of atomic hydrogen is approximately 39.492 × 10^7 nm.