Question #18492

1 Answer
Aug 7, 2017

#DeltaE_("vis") = (5.094 - 2.838) xx 10^(-19)color(white)(.)"J"#

#Deltanu_("vis") = (7.687 - 4.283) xx 10^(14)color(white)(.)"s"^(-1)" or Hz"#

for the visible light range of #390 - "700 nm"#.


Visible light is known to be in the region #390 - 700# #"nm"# (although I find it easier to just use #400 - 700#). We can start there.

The relationship between energy and frequency is given by the equation for energy #bbE# as quanta, #hnu# ("packets" of light, each one called a quantum of light):

#bb(E = hnu)#,

where #h = 6.626 xx 10^(-34) "J"cdot"s"# is Planck's constant, and #nu# is the frequency of the light in #"s"^(-1)#.

And the relationship between frequency #nu# and wavelength lambda is through the speed of light #c#:

#bb(nu = c/lambda)#,

with #lambda# in #"m"# and #c = 2.998 xx 10^(8)# #"m/s"#.

Or, combined, we get

#bb(E = (hc)/lambda)#.


#a)# The lower wavelength bound of #"390 nm"#, or #390 xx 10^(-9) "m"#, corresponds to an energy of:

#E = (hc)/lambda = ((6.626 xx 10^(-34) "J"cdot"s")(2.998 xx 10^(8) "m/s"))/(390 xx 10^(-9) "m")#

#= ul(5.094 xx 10^(-19)color(white)(.)"J")#

Likewise, the upper wavelength bound of #"700 nm"# will give you #E = ul(2.838 xx 10^(-19)color(white)(.)"J")#, and the range shall be:

#color(blue)(barul|stackrel(" ")(" "DeltaE_("vis") = (5.094 - 2.838) xx 10^(-19)color(white)(.)"J"" ")|)#

#b)# The lower wavelength bound of #"390 nm"#, or #390 xx 10^(-9) "m"#, corresponds to a frequency of:

#nu = E/h = c/lambda#

#= (2.998 xx 10^(8) "m/s")/(390 xx 10^(-9) "m")#

#= ul(7.687 xx 10^(14)color(white)(.)"s"^(-1)" or Hz")#

Likewise, the upper wavelength bound of #"700 nm"# will give you #E = ul(4.283 xx 10^(14)color(white)(.)"s"^(-1)" or Hz")#, and the range shall be:

#color(blue)(barul|stackrel(" ")(" "Deltanu_("vis") = (7.687 - 4.283) xx 10^(14)color(white)(.)"s"^(-1)" or Hz"" ")|)#