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Answer 1 · 2017-10-08T08:34:34

#E=2.29xx10^-19J#

Energy of photon = Planck's constant x frequency or #E=hf#.

Because #f# is not given, we use a different equation to find #f#. That is wave equation, which is #c=flambda#

where,

#c# is the speed of light/photon#= 3.00 xx 10^8 m.s#
#lambda# is the wavelength #(m)#
#f# is the frequency#(H_z)#

When we rearrange this formula to solve for #f#, we get #f=c/lambda#.

Keep in mind that wavelength is measured in metres, therefore we need to convert #=869nm to metres.

Nano is #10^-9#,
#869 xx 10^-9=8.69xx10^-7m#

#f=(3.00 xx 10^8)/(8.69 xx 10^-7)#
#= 3.45xx10^14H_z#

Now we are able to find #E.#

#=(6.63 xx 10^-34) xx ( 3.45 xx 10^14)#

#E=2.29xx10^-19J#