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Chapter 7: Problem 53

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electromagnetic radiation is emitted in each transition? a. \(n=3 \rightarrow n=2\) b. \(n=4 \rightarrow n=2\) c. \(n=2 \rightarrow n=1\)

Short Answer

Expert verified
a. The transition from \(n=3\) to \(n=2\) results in the emission of visible light with a wavelength of approximately \(656.1\) nm. b. The transition from \(n=4\) to \(n=2\) results in the emission of visible light with a wavelength of approximately \(486.1\) nm. c. The transition from \(n=2\) to \(n=1\) results in the emission of ultraviolet light with a wavelength of approximately \(121.5\) nm.

Step by step solution

01

Apply the Rydberg Formula

Calculate the wavelength using the Rydberg formula: \[\frac{1}{\lambda} = R_H\left(\frac{1}{2^2} - \frac{1}{3^2}\right)\]
02

Solve for the Wavelength

Obtain the wavelength (\(\lambda\)) and convert it to nanometers (nm): \[\lambda = \frac{1}{R_H\left(\frac{1}{2^2} - \frac{1}{3^2}\right)} \approx 656.1\, nm\]
03

Determine Electromagnetic Radiation Type

Since \(390 \leq \lambda \leq 750\) nm, the electromagnetic radiation emitted is in the visible light range. b. \(n = 4 \rightarrow n = 2\)
04

Apply the Rydberg Formula

Calculate the wavelength using the Rydberg formula: \[\frac{1}{\lambda} = R_H\left(\frac{1}{2^2} - \frac{1}{4^2}\right)\]
05

Solve for the Wavelength

Obtain the wavelength (\(\lambda\)) and convert it to nanometers (nm): \[\lambda = \frac{1}{R_H\left(\frac{1}{2^2} - \frac{1}{4^2}\right)} \approx 486.1\, nm\]
06

Determine Electromagnetic Radiation Type

Since \(390 \leq \lambda \leq 750\) nm, the electromagnetic radiation emitted is in the visible light range. c. \(n = 2 \rightarrow n = 1\)
07

Apply the Rydberg Formula

Calculate the wavelength using the Rydberg formula: \[\frac{1}{\lambda} = R_H\left(\frac{1}{1^2} - \frac{1}{2^2}\right)\]
08

Solve for the Wavelength

Obtain the wavelength (\(\lambda\)) and convert it to nanometers (nm): \[\lambda = \frac{1}{R_H\left(\frac{1}{1^2} - \frac{1}{2^2}\right)} \approx 121.5\, nm\]
09

Determine Electromagnetic Radiation Type

Since \(\lambda < 390\) nm, the electromagnetic radiation emitted is in the ultraviolet range.

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