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QuestionFind the wavelength of a photon with energy 6.89×10−19 J6.89 \times 10^{-19} \mathrm{~J}.

Studdy Solution

STEP 1

Assumptions1. The energy of the photon is 6.89×10−19 J6.89 \times10^{-19} \mathrm{~J} . We are to find the wavelength of the photon3. We will use the Planck-Einstein relation, which states that the energy of a photon is given by the product of Planck's constant (h) and the speed of light (c), divided by the wavelength of the photon (λ). The equation is =hcλ = \frac{hc}{\lambda}, where is the energy of the photon, h is Planck's constant (6.63×10−34 J⋅s6.63 \times10^{-34} \mathrm{~J \cdot s}), c is the speed of light (3.00×108 m/s3.00 \times10^{8} \mathrm{~m/s}), and λ is the wavelength of the photon.
4. We will rearrange the equation to solve for λ.

STEP 2

First, we need to rearrange the Planck-Einstein relation to solve for λ.
λ=hc\lambda = \frac{hc}{}

STEP 3

Now, plug in the given values for Planck's constant (h), the speed of light (c), and the energy of the photon () to calculate the wavelength.
λ=(6.63×10−34 J⋅s)×(3.00×108 m/s)6.89×10−19 J\lambda = \frac{(6.63 \times10^{-34} \mathrm{~J \cdot s}) \times (3.00 \times10^{8} \mathrm{~m/s})}{6.89 \times10^{-19} \mathrm{~J}}

STEP 4

Calculate the wavelength of the photon.
λ=(6.63×10−34 J⋅s)×(3.00×108 m/s)6.89×10−19 J=2.88×10−7 m\lambda = \frac{(6.63 \times10^{-34} \mathrm{~J \cdot s}) \times (3.00 \times10^{8} \mathrm{~m/s})}{6.89 \times10^{-19} \mathrm{~J}} =2.88 \times10^{-7} \mathrm{~m}The wavelength of the photon is 2.88×10−7 m2.88 \times10^{-7} \mathrm{~m}.

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