Question

1. what is the wavelength of a photon that has an energy of 3.4 x 10^{-19} j? identify the type of radiation.
2. calculate the frequency and energy of a photon with a wavelength of 1.18 x 10^{8} m.

Explanation:

Problem 6

Step1: Use the energy - wavelength formula

The energy of a photon is given by \(E = h
u=\frac{hc}{\lambda}\), where \(h = 6.626\times10^{-34}\space J\cdot s\) (Planck's constant), \(c=3\times 10^{8}\space m/s\) (speed of light), \(E\) is the energy of the photon, and \(\lambda\) is the wavelength. We can re - arrange the formula for \(\lambda\): \(\lambda=\frac{hc}{E}\)

Step2: Substitute the values

Substitute \(h = 6.626\times10^{-34}\space J\cdot s\), \(c = 3\times 10^{8}\space m/s\), and \(E=3.4\times 10^{-19}\space J\) into the formula:

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Step3: Identify the type of radiation

Visible light has a wavelength range of approximately \(400 - 700\space nm\) (\(4\times10^{-7}-7\times 10^{-7}\space m\)). Since \(\lambda = 5.846\times 10^{-7}\space m=584.6\space nm\), this photon is in the visible light range.

Step1: Calculate the frequency

We know that \(c=\lambda
u\), so \(
u=\frac{c}{\lambda}\). Substitute \(c = 3\times 10^{8}\space m/s\) and \(\lambda=1.18\times 10^{-8}\space m\)

$$ u=\frac{3\times 10^{8}}{1.18\times 10^{-8}}=2.542\times 10^{16}\space Hz $$

Step2: Calculate the energy

Use the formula \(E = h
u\). Substitute \(h = 6.626\times 10^{-34}\space J\cdot s\) and \(
u = 2.542\times 10^{16}\space Hz\)

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Answer:

The wavelength of the photon is \(\lambda = 5.85\times 10^{-7}\space m\) (rounded to three significant figures) and it is visible light.

Problem 7