An electron transitions between energy levels in an atom, releasing a photon with energy $3.2 \times 10^{-19} J$. We are asked to calculate the wavelength of this photon, given the speed of light $c = 3.0 \times 10^8 m/s$ and Planck's constant $h = 6.6 \times 10^{-34} J s$.

1. Problem Description

An electron transitions between energy levels in an atom, releasing a photon with energy $3.2 \times 10^{-19} J$. We are asked to calculate the wavelength of this photon, given the speed of light $c = 3.0 \times 10^8 m/s$ and Planck's constant $h = 6.6 \times 10^{-34} J s$.

2. Solution Steps

First, we need to use the formula that relates the energy of a photon to its wavelength:

$E = hf$

where $E$ is the energy of the photon, $h$ is Planck's constant, and $f$ is the frequency of the photon.

Also, the relationship between frequency and wavelength is:

$c = \lambda f$

where $c$ is the speed of light and $\lambda$ is the wavelength.

From this, we can say that $f = \frac{c}{\lambda}$.

Substituting $f = \frac{c}{\lambda}$ into the first equation, we get:

$E = h \frac{c}{\lambda}$

We can rearrange this equation to solve for the wavelength $\lambda$:

$\lambda = \frac{hc}{E}$

Now we can plug in the given values:

$\lambda = \frac{(6.6 \times 10^{-34} J s)(3.0 \times 10^8 m/s)}{3.2 \times 10^{-19} J}$

$\lambda = \frac{19.8 \times 10^{-26}}{3.2 \times 10^{-19}} m$

$\lambda = 6.1875 \times 10^{-7} m$

$\lambda = 618.75 \times 10^{-9} m$

$\lambda = 618.75 nm$

3. Final Answer

The wavelength of the photon is $6.1875 \times 10^{-7} m$ or $618.75 nm$.