Question #2d808

The thing to remember about the energy of a photon is that it is directly proportional to its frequency and, consequently, indirectly proportional to its wavelength.

In other words, the shorter the wavelength of the photon, the higher its frequency and the higher its energy.

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The energy of the photon can be calculated by using the Planck - Einstein equation

#color(blue)(ul(color(black)(E = h * nu)))#

Here

• #E# is the energy of the photon
• #nu# is the frequency of the photon
• #h# is Planck's constant, equal to #6.626 * 10^(-34)"J s"#

If you take into account the fact that

#color(blue)(ul(color(black)(nu * lamda = c)))#

Here

• #lamda# is the wavelength of the photon
• #c# is the speed of light in a vacuum, usually given as #3 * 10^8"m s"^(-1)#

you can say that the energy of the photon is equal to

#E = h * c/(lamda)#

Plug in your values to find the energy of the photon, but do not forget to convert the wavelength of the photon to meters!

#E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * (3 * 10^8 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(650 * 10^(-9)color(red)(cancel(color(black)("m"))))#

#E = color(darkgreen)(ul(color(black)(3.1 * 10^(-19)color(white)(.)"J")))#

To convert this to kilojoules per mole, use the fact that #1# mole of photons contains #6.022 * 10^(23)# photons.

The energy of #1# mole of photons will be equal to

#6.022 * 10^(23) color(red)(cancel(color(black)("photons"))) * (3.1 * 10^(-19)color(red)(cancel(color(black)("J"))))/(1color(red)(cancel(color(black)("photon")))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J"))))= "190 kJ"#

This means that the energy of the photons in kilojoules per mole is equal to

#E = color(darkgreen)(ul(color(black)("190 kJ mol"^(-1))))#

The answers are rounded to two sig figs, the number of sig figs you have for the wavelength of the photon.