A hypothetical electromagnetic wave is shown below. The frequency of the wave is $$x \times 10^{19}$$ Hz. $$x =$$ ______ (nearest integer)

1. Identify the Wavelength ($$\lambda$$)

Looking at the diagram, the distance of 1.5 pm represents the distance from a peak (crest) to a zero-crossing (node).

• A full wavelength ($$\lambda$$) consists of four such segments (peak to node, node to trough, trough to node, and node back to peak).
• Therefore, the wavelength is:
  $$\lambda = 4 \times 1.5 \text{ pm} = 6.0 \text{ pm}$$
• Converting picometers to meters:
  $$\lambda = 6.0 \times 10^{-12} \text{ m}$$

2. Relate Speed, Frequency, and Wavelength

Since this is an electromagnetic wave, it travels at the speed of light ($$c \approx 3 \times 10^8 \text{ m/s}$$). We use the standard wave equation:

$$c = \nu \lambda$$

Where:

• $$c$$ is the speed of light ($$3 \times 10^8 \text{ m/s}$$)
• $$\nu$$ is the frequency (Hz)
• $$\lambda$$ is the wavelength (m)

3. Calculate the Frequency ($$\nu$$)

Rearranging the formula to solve for frequency:

$$\nu = \frac{c}{\lambda} = \frac{3 \times 10^8 \text{ m/s}}{6 \times 10^{-12} \text{ m}}$$

Now, simplify the powers of ten:

$$\nu = 0.5 \times 10^{8 - (-12)} = 0.5 \times 10^{20} \text{ Hz}$$ 

$$\nu = 5 \times 10^{19} \text{ Hz}$$