Final answer:
The wavelength of the light emitted for an electronic transition from n=3 to n=2 in a hydrogen atom is approximately 656 nm, and the frequency is approximately 4.573 x 10^14 Hz.
Step-by-step explanation:
To calculate the frequency (in Hz) and wavelength (in nm) of light emitted for an electronic transition from n=3 to n=2 in a hydrogen atom, we can use the formula:
Rydberg equation:
1/λ = R(1/n1^2 - 1/n2^2)
Where λ is the wavelength, R is the Rydberg constant (1.097 x 10^7 m^-1), n1 is the initial energy level, and n2 is the final energy level.
Substituting the values into the equation:
1/λ = (1.097 x 10^7 m^-1)(1/3^2 - 1/2^2)
Simplifying the equation:
1/λ = (1.097 x 10^7 m^-1)(1/9 - 1/4)
Calculating:
1/λ = (1.097 x 10^7 m^-1)(5/36)
Solving for λ:
λ = 36/(1.097 x 10^7 m^-1)(5/36)
λ ≈ 656 nm
Therefore, the wavelength of the emitted light is approximately 656 nm.
To calculate the frequency, we can use the speed of light equation:
Speed of light equation:
c = λν
Where c is the speed of light (3 x 10^8 m/s), λ is the wavelength, and ν is the frequency.
Substituting the values into the equation:
3 x 10^8 m/s = (656 nm)(ν)
Converting the wavelength to meters:
3 x 10^8 m/s = (656 x 10^-9 m)(ν)
Simplifying the equation:
ν = (3 x 10^8 m/s)/(656 x 10^-9 m)
ν ≈ 4.573 x 10^14 Hz
Therefore, the frequency of the emitted light is approximately 4.573 x 10^14 Hz.