Question

Use the Balmer equation (8.2) to determine (a) the frequency, in \(s^{-1}\), of the radiation corresponding to \(n=5\) (b) the wavelength, in nanometers, of the line in the Balmer series corresponding to \(n=7\) (c) the value of \(n\) corresponding to the Balmer series line at \(380 \mathrm{nm}\)

Short answer

The calculated frequency for \(n=5\) is about \(6.166 \times 10^{14} s^{-1}\), the wavelength for \(n=7\) is approximately \(397 nm\) and the value of \(n\) for the Balmer series line at 380 nm is approximately 4.

Step-by-step solution

Calculation of frequency

Use the Balmer equation to find the wavelength for \( n=5 \), then use the formula \( v = \frac{c}{\lambda} \) to find the frequency. \( \lambda = \frac{1}{R_H (\frac{1}{2^2} - \frac{1}{5^2})} \) and \( v = \frac{c}{\lambda} \) Compute the values to get the frequency in \( s^{-1} \).

Calculation of wavelength

Again use the Balmer equation, this time to find the wavelength for \( n=7 \). Substitute the values to compute the wavelength in meters, and then convert to nanometers by multiplying the result by \( 10^9 \). That is, \( \lambda = \frac{1}{R_H (\frac{1}{2^2} - \frac{1}{7^2})} \times 10^9 nm \). Calculate this to get the wavelength in nanometers.

Finding the value of n

Rearrange the Balmer equation to solve for n. \( n = \sqrt{\frac{1}{(\frac{1}{\lambda} - \frac{1}{4R_H})}} \). Substitute the given wavelength at 380 nm (converting to meters by multiplying \( 10^{-9} \)) into the equation and solve to get the value of n.

Key concepts

Balmer equation

The Balmer equation is central to understanding the spectral lines of hydrogen. It specifically describes the lines in the visible spectrum and is crucial for calculating different properties of these lines. The equation is given by:

• \( \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \)

where \( \lambda \) is the wavelength of light, \( R_H \) is the Rydberg constant (1.097 x 10^7 \( m^{-1} \)), and \( n \) is the quantum number of the electron's energy level. This equation only applies to transitions ending in the second energy level \( (n=2) \), known as the Balmer series. It provides the foundation for calculating wavelength and leaps to higher quantum numbers.

frequency calculation

Frequency calculation involves determining how often a wave cycle passes a specific point per second when an electron in hydrogen transitions between energy levels. Once the wavelength \( \lambda \) is found using the Balmer equation, frequencies can be calculated using the equation:

• \( v = \frac{c}{\lambda} \)

where \( v \) is the frequency and \( c \) is the speed of light (approximately \( 3.00 \times 10^8 \, m/s \)).This method ensures a seamless conversion between the wavelength and frequency, both of which are essential parameters in characterizing light. For example, if you find \( \lambda \) for \( n=5 \), substitute back to compute \( v \), which reflects the frequency in cycles per second or hertz \( (Hz) \).

wavelength calculation

Wavelength calculation is pivotal in the study of atomic transitions, particularly in the Balmer series for hydrogen. Starting with the Balmer equation, you can compute the wavelength corresponding to a specific quantum number.When provided with the quantum number \( n \), place it into the formula:

• \( \lambda = \frac{1}{R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right)} \)

This value \( \lambda \) is initially in meters. To convert it to nanometers, which is more common in spectroscopy, multiply by \( 10^9 \). This conversion is straightforward and commonly required for practical applications, like when calculating \( n=7 \) Balmer series wavelengths.

quantum number calculation

Quantum number calculation involves determining the specific integer representing the energy level to which an electron transitions or falls within an atom.When a wavelength \( \lambda \) is given, for example 380 nm in the Balmer series, convert it to meters first by multiplying by \( 10^{-9} \).Use the rearranged form of the Balmer equation to find \( n \):

• \( n = \sqrt{\frac{1}{\left(\frac{1}{\lambda} - \frac{1}{4R_H}\right)}} \)

Substitute \( \lambda \) as required, navigate the algebra, and compute \( n \) using your result for \( \lambda \).This step requires precision, but it effectively bridges the physical to the mathematical understanding of spectral lines.