Question

1. calculate the wavelength of a photon with a frequency of 6.59 x 10^9 hz. identify the type of radiation.

c = λ·v
3.00 x 10^8 m/s = (λ)

1. a photon of light has an energy of 7.29 x 10^-15 j. calculate the wavelength & identify the type of radiation.

e = h·v
7.29 x 10^-15 j = (6.626 x 10^-34 j·s)(v)

1. a photon of light has an energy of 2.93 x 10^-24 j. calculate its frequency & identify the type of radiation.

2.93 x 10^-24 j

1. the wavelength of green light from a traffic signal is 520 nm. calculate the energy & frequency.

3.00 x 10^8 m/s = (520 x 10^-9 m)(v)

Explanation:

1. Calculate the wavelength of a photon with a frequency of \(6.59\times10^{9}\text{ Hz}\). Identify the type of radiation.

Step1: Use the formula \(c = \lambda\cdot

u\)
We know that the speed of light \(c = 3.00\times 10^{8}\text{ m/s}\), and the formula \(c=\lambda\cdot
u\) can be rearranged to \(\lambda=\frac{c}{
u}\).

Step2: Substitute the values

Substitute \(c = 3.00\times 10^{8}\text{ m/s}\) and \(
u=6.59\times 10^{9}\text{ Hz}\) (since \(1\text{ Hz}=1\text{ s}^{-1}\)) into the formula \(\lambda=\frac{3.00\times 10^{8}\text{ m/s}}{6.59\times 10^{9}\text{ s}^{-1}}\)

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This wavelength corresponds to microwave radiation (microwaves have wavelengths in the range of \(10^{-3}\text{ m}-1\text{ m}\))

Step1: Use the formula \(E = h\cdot

u\) to find \(
u\)
We know that Planck's constant \(h=6.626\times 10^{-34}\text{ J}\cdot\text{s}\). From \(E = h\cdot
u\), we can get \(
u=\frac{E}{h}\). Substitute \(E = 7.29\times 10^{-15}\text{ J}\) and \(h = 6.626\times 10^{-34}\text{ J}\cdot\text{s}\)

$$ u=\frac{7.29\times 10^{-15}\text{ J}}{6.626\times 10^{-34}\text{ J}\cdot\text{s}}\approx1.10\times 10^{19}\text{ s}^{-1} $$

Step2: Use \(c=\lambda\cdot

u\) to find \(\lambda\)
Since \(c = 3.00\times 10^{8}\text{ m/s}\) and \(
u = 1.10\times 10^{19}\text{ s}^{-1}\), from \(\lambda=\frac{c}{
u}\)

$$ \lambda=\frac{3.00\times 10^{8}\text{ m/s}}{1.10\times 10^{19}\text{ s}^{-1}}\approx2.73\times 10^{-11}\text{ m} $$

This wavelength corresponds to X - ray radiation (X - rays have wavelengths in the range of \(10^{-11}\text{ m}-10^{-8}\text{ m}\))

Step1: Use the formula \(E = h\cdot

u\)
From \(E = h\cdot
u\), we can solve for \(
u\) as \(
u=\frac{E}{h}\). Substitute \(E = 2.93\times 10^{-24}\text{ J}\) and \(h=6.626\times 10^{-34}\text{ J}\cdot\text{s}\)

$$ u=\frac{2.93\times 10^{-24}\text{ J}}{6.626\times 10^{-34}\text{ J}\cdot\text{s}}\approx4.42\times 10^{9}\text{ s}^{-1} $$

This frequency corresponds to radio - wave radiation (radio waves have frequencies in the range of \(10^{4}\text{ Hz}-10^{9}\text{ Hz}\))

Answer:

The wavelength \(\lambda = 4.55\times 10^{-2}\text{ m}\) and the radiation type is microwave.

2. A photon of light has an energy of \(7.29\times 10^{-15}\text{ J}\). Calculate the wavelength & identify the type of radiation.