Question

1. a wave has a frequency of 8.0×10^6 hz. calculate the wavelength. 12. what is the energy of a photon with a frequency of 6.0×10^14 hz?

Explanation:

Step1: Recall the formula for wavelength

The formula for the speed of a wave is \(c = \lambda f\), where \(c\) is the speed of light (\(c=3\times 10^{8}\space m/s\)), \(\lambda\) is the wavelength, and \(f\) is the frequency. Rearranging for \(\lambda\) gives \(\lambda=\frac{c}{f}\).

Step2: Substitute the values

Given \(f = 8.0\times 10^{6}\space Hz\) and \(c = 3\times 10^{8}\space m/s\). Substitute into the formula: \(\lambda=\frac{3\times 10^{8}}{8.0\times 10^{6}}\).

Step3: Calculate the wavelength

\(\lambda=\frac{3\times 10^{8}}{8.0\times 10^{6}}=\frac{3}{8.0}\times10^{8 - 6}= 0.375\times10^{2}=37.5\space m\)

Step4: Recall the formula for photon energy

The energy of a photon is given by \(E = hf\), where \(h\) is Planck's constant (\(h = 6.63\times 10^{-34}\space J\cdot s\)) and \(f\) is the frequency.

Step5: Substitute the values

Given \(f=6.0\times 10^{14}\space Hz\). Substitute into the formula: \(E=(6.63\times 10^{-34})\times(6.0\times 10^{14})\).

Step6: Calculate the photon energy

\(E=(6.63\times6.0)\times10^{-34 + 14}=39.78\times10^{-20}=3.978\times 10^{-19}\space J\)

Answer:

1. The wavelength is \(37.5\space m\).
2. The energy of the photon is \(3.978\times 10^{-19}\space J\)