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Question #ae5a5

1 Answer

Nov 5, 2017

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

Explanation:

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

$E = 3.38 \times 10^{- 19}$ $E = 3.38xx10^-19$ J

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

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