The energy of electron in the \(\mathrm{n}^{\text {th }}\) orbit of hydrogen atom is expressed as \(E_{n}=-\left[(13.6) / \mathrm{n}^{2}\right] e v .\) The shortest and longest wave length of lyman series will be. (A) \(912 \bar{\AA}, 1216 \AA\) (B) \(1315 \AA, 1530 \AA\) (C) \(5463 \AA, 7858 \AA\) (D) None of these

For the shortest and longest wavelength in Lyman series, we consider the transitions from orbit n = 2 to n = 1 (longest wavelength) and n = ∞ to n = 1 (shortest wavelength). Let's find the change in energy for these transitions: For n = 2 to n = 1: ΔE₁ = |E₂ - E₁| = \(\left|\frac{-13.6}{2^2} - \frac{-13.6}{1^2}\right|\) eV For n = ∞ to n = 1: ΔE₂ = |E_∞ - E₁| = \(\left|\frac{-13.6}{\infty^2} - \frac{-13.6}{1^2}\right|\) eV

Next, we need to convert the energy differences ΔE₁ and ΔE₂ to wavelengths using the formula: ΔE = h\(\frac{c}{\lambda}\), where ΔE is the change in energy, h is Planck's constant (4.136 × 10⁻¹⁵ eV·s), and c is the speed of light (3 × 10⁸ m/s). Longest wavelength (n = 2 to n = 1): \(\lambda_1 = \frac{hc}{\Delta E_1}\) Shortest wavelength (n = ∞ to n = 1): \(\lambda_2 = \frac{hc}{\Delta E_2}\)

Calculate the shortest and longest wavelengths using the energy differences obtained in step 1: Longest wavelength (n = 2 to n = 1): \(\lambda_1 = \frac{(4.136 \times 10^{-15})(3 \times 10^8)}{\Delta E_1}\) Shortest wavelength (n = ∞ to n = 1): \(\lambda_2 = \frac{(4.136 \times 10^{-15})(3 \times 10^8)}{\Delta E_2}\) Finally, convert these wavelengths from meters to Ångströms by multiplying by \(10^{10}\): \(\lambda_1 \times 10^{10} = \) Longest wavelength in Ångströms \(\lambda_2 \times 10^{10} = \) Shortest wavelength in Ångströms After calculating the values, we find that the shortest and longest wavelengths of the Lyman series are approximately \(1216 \mathring{A}\) and \(912 \mathring{A}\), respectively. This corresponds to the answer choice (A) \(912 \mathring{A}, 1216 \mathring{A}\).