The problem asks to calculate the wavelength of a photon released when an electron jumps from one energy level to another, given the energy released and the values of Planck's constant ($h$) and the speed of light ($c$). The energy released is $3.2 \times 10^{-19} J$, Planck's constant is $6.6 \times 10^{-34} J s$, and the speed of light is $3.0 \times 10^8 m/s$.

1. Problem Description

The problem asks to calculate the wavelength of a photon released when an electron jumps from one energy level to another, given the energy released and the values of Planck's constant ($h$) and the speed of light ($c$). The energy released is $3.2 \times 10^{-19} J$, Planck's constant is $6.6 \times 10^{-34} J s$, and the speed of light is $3.0 \times 10^8 m/s$.

2. Solution Steps

We can 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.

We also know that the speed of light ($c$) is related to the frequency ($f$) and wavelength ($\lambda$) of the photon by the formula:

$c = f\lambda$

From this, we can express the frequency as:

$f = \frac{c}{\lambda}$

Substituting this expression for $f$ into the first equation, we get:

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

Now, we can solve for the wavelength $\lambda$:

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

Plugging in the given values:

$h = 6.6 \times 10^{-34} J s$

$c = 3.0 \times 10^8 m/s$

$E = 3.2 \times 10^{-19} J$

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

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

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

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

$\lambda = 618.75 nm$

3. Final Answer

The wavelength of the photon is $618.75 \times 10^{-9} m$ or $618.75 nm$.