# Question #1427c

Nov 16, 2017

${E}_{\gamma} = 4.31 \cdot {10}^{- 33} J = 26.94$ $f e V$

#### Explanation:

${E}_{\gamma} = h v$
where:

• ${E}_{\gamma}$ is the photon energy (in $J$).
• $h$ is the Planck constant ($6.63 \cdot {10}^{- 34} J \cdot s$).
• $v$ is the frequency (in $H z = \frac{1}{s}$).

So if you substitute with the values,
${E}_{\gamma} = 6.63 \cdot {10}^{- 34} \cdot 6.5$
${E}_{\gamma} = 4.31 \cdot {10}^{- 33} J$

if you want it in $e V$ and not in $J$, you use the equivalence:
$1$ $e V = 1.6 \cdot {10}^{- 19}$ $J$

so,
${E}_{\gamma} = 2.694 \cdot {10}^{- 14} e V$

or,
${E}_{\gamma} = 26.94$ $f e V$