# The wavelength of a photon in nm having energy of 3.083•10^12J is approximately?

$6.4515 \cdot {10}^{-} 29 n m$
The energy of a photon is the frequency multiplied by the Planck constant; $E = h f$, and the wavelength of a wave is its speed divided by it's frequency; $v = f \lambda \implies f = \setminus \frac{v}{\lambda}$, so it's energy, $E = \setminus \frac{h v}{\lambda}$. This rearranges to $\lambda = \setminus \frac{h v}{E}$, where $h$ is the Planck constant, $v$ is the speed of light, and $E$ is the energy of the photon $\therefore \lambda = \setminus \frac{6.63 \cdot {10}^{- 34} \cdot 3.0 \cdot {10}^{8}}{3.083 \cdot {10}^{12}} = 6.4515 \cdot {10}^{-} 38 m$
$= 6.4515 \cdot {10}^{-} 29 n m$
I guess that you meant to write $3.083 \cdot {10}^{- 12}$ not $3.083 \cdot {10}^{12}$, seeing as gamma rays have a wavelength of ~10^-11m, which is ~10^-18 times larger.
If $E = 3.083 \cdot {10}^{-} 12 J$, using the same method, it would give a wavelength of $6.4515 \cdot {10}^{-} 5 n m$, which is a reasonable value for electromagnetic waves.