This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Morning hafsatjibril2006 — let's get this done. To find the wavelength of the second line in the Balmer series of the hydrogen spectrum, we use the Rydberg formula: (1)/() = R ( (1)/(n_f^2) - (1)/(n_i^2) ) where: • is the wavelength • R is the Rydberg constant for hydrogen (1.097 × 10^-2 nm^-1) • n_f is the final principal quantum number • n_i is the initial principal quantum number Step 1: Identify the given values. For the Balmer series, the electron transitions to the n_f = 2 energy level. The "second line" in the Balmer series corresponds to a transition from n_i = 4 to n_f = 2. (The first line is n_i=3 n_f=2, the second is n_i=4 n_f=2, etc.) The Rydberg constant R = 1.097 × 10^-2 nm^-1. Step 2: Substitute the values into the Rydberg formula. (1)/() = (1.097 × 10^-2 nm^-1) ( (1)/(2^2) - (1)/(4^2) ) Step 3: Calculate the terms inside the parenthesis. (1)/() = (1.097 × 10^-2 nm^-1) ( (1)/(4) - (1)/(16) ) (1)/() = (1.097 × 10^-2 nm^-1) ( (4)/(16) - (1)/(16) ) (1)/() = (1.097 × 10^-2 nm^-1) ( (3)/(16) ) (1)/() = (1.097 × 10^-2 nm^-1) (0.1875) Step 4: Multiply by the Rydberg constant. (1)/() = 0.002056875 nm^-1 Step 5: Calculate the wavelength . = (1)/(0.002056875 nm)^-1 ≈ 486.19 nm Comparing this value to the given options, 486.19 nm is closest to 486.3 nm. The final answer is c. 486.3nm. Got more? Send 'em.