#E=h*nu#,
where:
#h# is Planck's constant and #nu# is frequency.
We can determine the frequency from the relationship between the speed of light, wavelength, and frequency.
#c=lambdaxxnu#,
where:
#c# is the speed of light, #lambda# (pronounced lamda) is wavelength, and #nu# (pronounced noo) is frequency.
We can rearrange the equation to isolate #nu#.
#nu=c/lambda#
We can substitute #c/lambda# for #nu# into the equation for the energy of a photon. We will need to convert wavelength in nm to m, since the speed of light is #3.00xx10^8# #"m/s"#.
Known
#h=6.626xx10^(−34)# #"J"*"s"#
#lambda=350color(red)cancel(color(black)("nm"))xx("1 m")/(1xx10^9color(red)cancel(color(black)("nm")))=3.5xx10^(-7)# #"m"#.
#c=3.00xx10^8# #"m/s"# to three significant figures
#nu=c/lambda#
Unknown
#E#
Solution
Plug in the known values and solve.
#E=(6.626xx10^(−34)"J"*color(red)cancel(color(black)("s")))xx(3.00xx10^8color(red)cancel(color(black)("m"))/color(red)cancel(color(black)("s")))/(3.5xx10^(-7)color(red)cancel(color(black)("m")))=5.7xx10^-19# #"J"# (rounded to two significant figures)
If you prefer to calculate #nu# separately, divide #c# by #lambda#.
#nu=c/lambda=(3.00xx10^8color(red)cancel(color(black)("m"))/"s")/(3.5xx10^(-7)color(red)cancel(color(black)("m")))=#
#(8.6xx10^14)/"s"=8.6xx10^14# #"Hz"#
#E=6.626xx10^(-34)"J"*color(red)cancel(color(black)("s"))xx(8.6xx10^14)/color(red)cancel(color(black)("s"))=5.7xx10^(-19)# #"J"# (rounded to two significant figures)
