91影视

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

Next, we need to convert the energy differences 螖E鈧� and 螖E鈧� to wavelengths using the formula: 螖E = h\(\frac{c}{\lambda}\), where 螖E is the change in energy, h is Planck's constant (4.136 脳 10鈦宦光伒 eV路s), and c is the speed of light (3 脳 10鈦� m/s). Longest wavelength (n = 2 to n = 1): \(\lambda_1 = \frac{hc}{\Delta E_1}\) Shortest wavelength (n = 鈭� 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 = 鈭� 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 脜ngstr枚ms by multiplying by \(10^{10}\): \(\lambda_1 \times 10^{10} = \) Longest wavelength in 脜ngstr枚ms \(\lambda_2 \times 10^{10} = \) Shortest wavelength in 脜ngstr枚ms 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) \(912 \mathring{A}, 1216 \mathring{A}\).