In a hydrogen atom, what is the wavelength of the emitted photon when an electron transitions from n=3 to n=2?

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Q: In a hydrogen atom, what is the wavelength of the emitted photon when an electron transitions from n=3 to n=2?

Solution: Using the Rydberg formula, the wavelength for the transition from n=3 to n=2 is approximately 486 nm.

Steps: 8

Step 1: Understand that in a hydrogen atom, electrons can move between different energy levels, which are represented by 'n' values (like n=1, n=2, n=3, etc.).
Step 2: Identify the initial and final energy levels for the transition. Here, the electron is moving from n=3 (higher energy level) to n=2 (lower energy level).
Step 3: Use the Rydberg formula to calculate the wavelength of the emitted photon. The formula is: 1/λ = R * (1/n1^2 - 1/n2^2), where R is the Rydberg constant (approximately 1.097 x 10^7 m^-1), n1 is the lower energy level (n=2), and n2 is the higher energy level (n=3).
Step 4: Plug in the values into the formula: 1/λ = 1.097 x 10^7 m^-1 * (1/2^2 - 1/3^2).
Step 5: Calculate the values inside the parentheses: 1/2^2 = 1/4 = 0.25 and 1/3^2 = 1/9 ≈ 0.1111. So, 0.25 - 0.1111 = 0.1389.
Step 6: Now, multiply this result by the Rydberg constant: 1/λ = 1.097 x 10^7 m^-1 * 0.1389.
Step 7: Calculate 1/λ to find λ: 1/λ ≈ 1.527 x 10^6 m^-1, so λ = 1 / (1.527 x 10^6) m.
Step 8: Convert the wavelength from meters to nanometers (1 m = 1 x 10^9 nm): λ ≈ 486 nm.