Question

Question #45665

Answer 1

`E approx 4513 \ eV`

Explanation:

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`