Question

The water molecule has an \(l\) = 1 rotational level 1.01 \(\times\) 10\(^-$$^5\) eV above the \(l\) = 0 ground level. Calculate the wavelength and frequency of the photon absorbed by water when it undergoes a rotational-level transition from \(l\) = 0 to \(l\) = 1. The magnetron oscillator in a microwave oven generates microwaves with a frequency of 2450 MHz. Does this make sense, in view of the frequency you calculated in this problem? Explain.

Short answer

The photon frequency is approximately 24.4 GHz and exceeds the microwave's 2.45 GHz by a factor of 10, indicating a higher frequency.

Step-by-step solution

Calculate the Energy Difference

The energy difference between the two rotational levels is given as the energy of the rotatational level when \( l = 1 \), which is \( 1.01 \times 10^{-5} \) eV.

Convert Energy to Joules

To convert the energy from electron volts (eV) to joules (J), use the conversion factor: \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \). Therefore, the energy difference in joules is: \[ E = 1.01 \times 10^{-5} \, \text{eV} \times 1.602 \times 10^{-19} \, \frac{\text{J}}{\text{eV}} = 1.618 \times 10^{-24} \, \text{J}.\]

Relate Energy to Wavelength

Use Planck's relation \( E = \frac{hc}{\lambda} \) to find the wavelength \( \lambda \), where \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \) and \( c = 3 \times 10^8 \, \text{m/s} \). Solve for \( \lambda \): \[ \lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \times 3 \times 10^8 \, \text{m/s}}{1.618 \times 10^{-24} \, \text{J}} \approx 1.23 \times 10^{-2} \, \text{m}. \]

Calculate Frequency of the Photon

From the wavelength \( \lambda \), the frequency \( u \) can be found using \( c = \lambda u \). Rearranging gives: \[ u = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \text{m/s}}{1.23 \times 10^{-2} \, \text{m}} \approx 2.44 \times 10^{10} \, \text{Hz}. \]

Compare to Microwave Frequency

Convert the microwave frequency from MHz to Hz: 2450 MHz = \( 2.45 \times 10^9 \, \text{Hz} \). The photon frequency \( 2.44 \times 10^{10} \, \text{Hz} \) we calculated is about 10 times higher than the frequency of microwaves generated by the oven's magnetron.

Key concepts

Energy Conversion

In many scientific contexts, energy values are often given in electron volts (eV), but it can be helpful to convert these values into joules (J) for more widespread applications. An electron volt is a unit of energy that represents the kinetic energy gained by an electron when it accelerates through an electric potential difference of one volt. To convert energy from eV to joules, the conversion factor used is that 1 eV equals approximately \( 1.602 \times 10^{-19} \) J.
For example, if we have an energy difference between rotational levels of \( 1.01 \times 10^{-5} \) eV, this can be converted to joules by multiplying with the conversion factor, resulting in \( 1.618 \times 10^{-24} \) J. This conversion is crucial because it allows us to make use of joules in calculations involving Planck's constant and other fundamental constants, which are generally expressed in joules.
Understanding how to make these conversions enables the use of different scientific principles to solve problems involving energy transitions, such as those encountered in rotational spectra.

Frequency Calculation

Frequency is an important property of waves, including electromagnetic waves like light. It is defined as the number of wave cycles that pass a point per second and is measured in hertz (Hz). In order to determine the frequency of a photon absorbed during a rotational-level transition, such as from \( l = 0 \) to \( l = 1 \), we must first find the wavelength.
Using Planck's relation \( E = \frac{hc}{\lambda} \) where \( E \) is energy, \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \), and \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \), we can rearrange this equation to solve for frequency \( u \) once we know the wavelength \( \lambda \). We find that \( u = \frac{c}{\lambda} \).
For this particular problem, if we calculate \( \lambda \) to be approximately \( 1.23 \times 10^{-2} \, \text{m} \), the frequency \( u \) of the photon is about \( 2.44 \times 10^{10} \) Hz. Frequency calculation gives us deeper insight into the properties of light and the types of transitions it can induce.

Wavelength Determination

The wavelength of a photon involved in a transition can be determined using its relationship with energy and frequency. In the equation \( E = \frac{hc}{\lambda} \), wavelength \( \lambda \) can be isolated to find \( \lambda = \frac{hc}{E} \).
Using the constants \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \) and \( c = 3 \times 10^8 \, \text{m/s} \), we can calculate the wavelength given an energy change of \( 1.618 \times 10^{-24} \) J. This results in a wavelength of approximately \( 1.23 \times 10^{-2} \, \text{m} \).
Wavelength determines the type of electromagnetic radiation; in this case, it falls in the microwave region of the spectrum. Knowing the wavelength helps in identifying where these transitions occur and can be essential in applications such as spectroscopy and communication technologies.

Microwave Frequency Comparison

When comparing frequencies, such as the calculated photon frequency and that of a microwave oven, we gain insight into the nature of electronic transitions. Here, we calculated the frequency of the photon absorbed by water during a rotational-level transition as approximately \( 2.44 \times 10^{10} \) Hz. In contrast, the microwave oven frequency is given as \( 2450 \) MHz, which converts to \( 2.45 \times 10^9 \) Hz.
This comparison highlights that the photon frequency is about 10 times higher than the frequency found in a typical microwave oven.

• The microwave oven's frequency is used for heating food through microwave radiation, which causes polar molecules to rotate, generating heat.
• Our calculated frequency corresponds to a different range of the electromagnetic spectrum, implicating different uses and mechanisms.

Understanding such distinctions is crucial in designing devices that interact with specific parts of the electromagnetic spectrum, further showing the interdisciplinary nature of physics, chemistry, and engineering.