Answer
For transitions n=3 to n=2, and n=4 to n=2, the emitted light is in the visible range; while for the transition n=2 to n=1, it is in the ultraviolet range.
Work Step by Step
【Tips】The question involves calculating the wavelength of light emitted during electron transitions in a hydrogen atom. The solution involves using the Rydberg formula which incorporates the Rydberg constant and quantum numbers of the initial and final energy levels. Depending on the wavelengths obtained, we can categorize the emitted radiation into different regions of the electromagnetic spectrum.
【Description】
Let's calculate the wavelengths for the transitions mentioned in the question:
a. n=3 to n=2:
Using the Rydberg formula,
λ = 1/R [(1/(n_lower)^2 - (1/(n_higher)^2)]
where R is the Rydberg constant (1.097 x 10^7 m^-1), n_lower = 2, and n_higher = 3. Substituting these values into the equation, we find λ = 1.9 x 10^-7 m or 656.3 nm. This wavelength falls within the visible light range.
b. n=4 to n=2:
Again, using the Rydberg formula and substituting the respective values (n_lower = 2, n_higher = 4), we get λ = 1.21 x 10^-7 m or 486.1 nm. This too falls within the visible light range.
c. n=2 to n=1:
Using the Rydberg formula for this transition (n_lower = 1, n_higher = 2), we find λ = 1.22 x10^-7 m or 121.6 nm. This falls within the ultraviolet light range.
【Final Answer】
For transitions n=3 to n=2, and n=4 to n=2, the emitted light is in the visible range; while for the transition n=2 to n=1, it is in the ultraviolet range.
