# Question #ae5a5

##### 1 Answer
Nov 5, 2017

Because $E = h . f$ and $f = \frac{c}{\lambda}$ we can solve this.

#### Explanation:

$E = \frac{h . c}{\lambda}$ and $\lambda$ is the only variable ...

$E = 3.38 \times {10}^{-} 19$J

Converting to eV (dividing by the charge on an electron) gives E = 2.11 eV