Question #1490f
1 Answer
Explanation:
Start by converting the wavelength of the photons from angstrom to meters by using the fact that
#1color(white)(.)stackrel(@)("A") = 1 * 10^(-10)# #"m"#
You should end up with
#7000 color(red)(cancel(color(black)(stackrel(@)("A")))) * (1 * 10^(-10)color(white)(.)"m")/(1color(red)(cancel(color(black)(stackrel(@)("A"))))) = 7 * 10^(-7)# #"m"#
Next, calculate the frequency of a single photon by using the fact that frequency and wavelength have an inverse relationship described by the equation
#color(blue)(ul(color(black)(nu * lamda = c)))#
Here
#lamda# is the wavelength of the photon#nu# is the frequency of the photon#c# is the speed of light in a vacuum, usually given as#3 * 10^8"m s"^(-1)#
Rearrange to solve for the frequency of the photon
#nu * lamda = c implies nu = c/(lamda)#
Plug in your value to find
#nu = (3 * 10^8 color(red)(cancel(color(black)("m"))) "s"^(-1))/(7 * 10^(-7)color(red)(cancel(color(black)("m")))) = 4.3 * 10^(14)# #"s"^(-1)#
Now, you know that the energy of a photon is directly proportional to its frequency, i.e. the higher the frequency, the higher the energy, as described by the Planck - Einstein equation
#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"#
Use this equation to calculate the energy of a single photon of this frequency
#E = 6.626 * 10^(-34)color(white)(.)"J" color(red)(cancel(color(black)("s"))) * 4.3 * 10^(14)color(red)(cancel(color(black)("s"^(-1))))#
#E = 2.85 * 10^(-19)# #"J"#
Since you know that the total energy provided by the photons must be equal to
#1 color(red)(cancel(color(black)("J"))) * "1 photon"/(2.85 * 10^(-19)color(red)(cancel(color(black)("J")))) = color(darkgreen)(ul(color(black)(4 * 10^(18)color(white)(.)"photons")))#
The answer is rounded to one significant figure, the number of sig figs you have for your values.
