which of the following series in the spectrum of hydrogen atom lies in the visible legion of the electro magnetic spectrum? (A) Paschen (B) Lyman (C) Brakett (D) Balmer

We can use the Rydberg formula to find the wavelength of transitions in the hydrogen spectrum: \(\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\) where λ is the wavelength, R_H is the Rydberg constant for hydrogen (\(1.097 \times 10^7 m^{-1}\)), n_1 and n_2 are the initial and final energy states of the electron (with \(n_1 < n_2\)), respectively.

We will calculate the wavelength for each series by substituting the n values. (A) Paschen series - Transitions from n ≥ 4 to n = 3: \(\frac{1}{\lambda_{Paschen}} = R_H \left(\frac{1}{3^2} - \frac{1}{4^2} \right)\) The shortest wavelength is when the electron transitions from n = 4 to n = 3 which corresponds to the highest energy transition. Any other transition will have a longer wavelength. (B) Lyman series - Transitions from n ≥ 2 to n = 1: \(\frac{1}{\lambda_{Lyman}} = R_H \left(\frac{1}{1^2} - \frac{1}{2^2} \right)\) The shortest wavelength is when the electron transitions from n = 2 to n = 1 which corresponds to the highest energy transition. Any other transition will have a longer wavelength. (C) Brackett series - Transitions from n ≥ 5 to n = 4: \(\frac{1}{\lambda_{Brackett}} = R_H \left(\frac{1}{4^2} - \frac{1}{5^2} \right)\) The shortest wavelength is when the electron transitions from n = 5 to n = 4, which corresponds to the highest energy transition. Any other transition will have a longer wavelength. (D) Balmer series - Transitions from n ≥ 3 to n = 2: \(\frac{1}{\lambda_{Balmer}} = R_H \left(\frac{1}{2^2} - \frac{1}{3^2} \right)\) The shortest wavelength is when the electron transitions from n = 3 to n = 2, which corresponds to the highest energy transition. Any other transition will have a longer wavelength.

Calculate the smallest wavelength for each series and compare it to the bounds of the visible spectrum (400 nm to 700 nm): (A) Paschen series: \(\lambda_{Paschen} = \frac{1}{1.097 \times 10^7(1/9 - 1/16)} = 8.21 \times 10^{-7}\, m\) or 821 nm, which falls outside the visible region. (B) Lyman series: \(\lambda_{Lyman} = \frac{1}{1.097 \times 10^7(1 - 1/4)} = 1.215 \times 10^{-7}\, m\) or 121.5 nm, which falls outside the visible region. (C) Brackett series: \(\lambda_{Brackett} = \frac{1}{1.097 \times 10^7(1/16 - 1/25)} = 1.45 \times 10^{-6}\, m\) or 1450 nm, which falls outside the visible region. (D) Balmer series: \(\lambda_{Balmer} = \frac{1}{1.097 \times 10^7(1/4 - 1/9)} = 6.563 \times 10^{-7}\, m\) or 656.3 nm, which falls within the visible region. Since the shortest wavelength of the Balmer series falls within the visible region of the electromagnetic spectrum, the correct answer is (D) Balmer series.