Question #45665

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Answer 1 · 2017-03-15T19:39:29

$E approx 4513 \ eV$

The energy of a photon is:

$E = hf = (hc)/lambda$ , as speed $c = f lambda$

....and where Plank's constant is $h approx 6.63 times 10^(-34) m^2 kg s^(-1)$

So we have:

$E = (6.63 times 10^(-34) cdot 3 times 10^8)/(0.275 times 10^(-9)$

$implies E approx 7.23 times 10^(-16) \ J$

And as: $1 \ J approx 6.24 times 10^(18) eV$

$E approx 4513 \ eV$