What is the wavelength of light emitted when an electron transitions from n=3 to n=2 in a hydrogen atom?

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Q: What is the wavelength of light emitted when an electron transitions from n=3 to n=2 in a hydrogen atom?

Solution: Using the Rydberg formula, the wavelength for the transition from n=3 to n=2 is approximately 434 nm.

Steps: 8

Step 1: Understand that in a hydrogen atom, electrons can exist in different energy levels, which are represented by 'n' values. Here, n=3 is a higher energy level and n=2 is a lower energy level.
Step 2: When an electron moves from a higher energy level (n=3) to a lower energy level (n=2), it emits light. The energy of the emitted light corresponds to the difference in energy between these two levels.
Step 3: Use the Rydberg formula to calculate the wavelength of the emitted light. 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 (2), and n2 is the higher energy level (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: 1/λ = 1.097 x 10^7 m^-1 * (1/4 - 1/9).
Step 6: Simplify the expression: 1/λ = 1.097 x 10^7 m^-1 * (0.25 - 0.1111) = 1.097 x 10^7 m^-1 * 0.1389.
Step 7: Calculate the result to find λ: λ = 1 / (1.097 x 10^7 m^-1 * 0.1389).
Step 8: Convert the wavelength from meters to nanometers (1 nm = 1 x 10^-9 m) to get the final answer, which is approximately 434 nm.