Question #ae5a5

1 Answer
Mark C.
Nov 5, 2017

Because #E = h.f# and #f = c/lambda# we can solve this.

Explanation:

#E = (h.c)/lambda # and #lambda# is the only variable ...

#E = 3.38xx10^-19#J

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

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