Question #ef39d
1 Answer
Explanation:
The first thing you need to do here is to figure out the frequency of the gamma-ray photons.
As you know, wavelength and frequency have an inverse relationship as given by the equation
#color(blue)(ul(color(black)(nu * lamda = c)))#
Here
#lamda# is the wavelength of the wave#c# is the speed of light in a vacuum, usually given as#3 * 10^8"m s"^(-1)#
Rearrange the above equation to solve for
#nu * lamda = c implies nu = c/(lamda)#
In your case, a wavelength of
#2.43 * 10^(-5) color(red)(cancel(color(black)("nm"))) * "1 m"/(10^9color(red)(cancel(color(black)("nm")))) = 2.43 * 10^(-14)"m"#
will correspond to a frequency of
#nu = (3 * 10^8 color(red)(cancel(color(black)("m")))"s"^(-1))/(2.43 * 10^(-14)color(red)(cancel(color(black)("m")))) = 1.2346 * 10^(22)"s"^(-1)#
Now, to find the energy of a single gamma ray photon, use the Planck - Einstein relation
#color(blue)(ul(color(black)(E = h * nu)))#
Here
#E# is the energy of the photon#h# is Planck's constant, equal to#6.626 * 10^(-34)"J s"# #nu# is the frequency of the photon
In your case, you will have
#E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 1.2346 * 10^(22)color(red)(cancel(color(black)("s"^(-1))))#
#E = 8.180 * 10^(-12)"J"#
In order to find the energy of one mole of gamma ray photons, use the fact that
#color(blue)(ul(color(black)("1 mole photons" = 6.022 * 10^(23)color(white)(.)"photons")))#
In your case, you will have
#6.022 * 10^(23) color(red)(cancel(color(black)("photons"))) * (8.180 * 10^(-12)"J")/(1color(red)(cancel(color(black)("photon")))) = 4.53 * 10^(12)"J"#
Since this represents the energy of
#"energy" = color(darkgreen)(ul(color(black)(4.53 * 10^(12)"J mol"^(-1))))#
A more common unit to use here is actually kilojoules per mole
#"energy" = color(darkgreen)(ul(color(black)(4.53 * 10^(9)"kJ mol"^(-1))))#
The answer is rounded to three sig figs.
