Question #2e3c2
1 Answer
Here's what I got.
Explanation:
Actually, all you have to do here is convert the frequency of the photon to wavelength by using the equation
#color(blue)(ul(color(black)(lamda * nu = c)))#
Here
#lamda# is the wavelength of the photon#nu# is its frequency#c# is the speed of light in a vacuum, usually given as#3.00 * 10^8"m s"^(-1)#
Keep in mind that you have
#"1 MHz" = 10^6"Hz" " "# and#" ""1 Hz" = "1 s"^(-1)#
which means that the frequency of the photon is equal to
#0.15 color(red)(cancel(color(black)("MHz"))) * (10^6color(red)(cancel(color(black)("Hz"))))/(1color(red)(cancel(color(black)("MHz")))) * "1 s"^(-1)/(1color(red)(cancel(color(black)("Hz")))) = 1.5 * 10^5"s"^(-1)#
Rearrange the equation to solve for
#lamda * nu = c implies lamda = c/(nu)#
Plug in your value to get
#lamda = (3.00 * 10^8 "m" color(red)(cancel(color(black)("s"^(-1)))))/(1.5 * 10^6color(red)(cancel(color(black)("s"^(-1))))) = color(darkgreen)(ul(color(black)(2.0 * 10^2"m")))#
The answer is rounded to two sig figs, the number of sig figs you have for the frequency of the photon.
Now, the second equation you listed
#color(blue)(ul(color(black)(E = h * nu)))#
where
#E# is the energy of the photon#h# is Planck's constant, equal to#6.626 * 10^(-34)"J s"#
is called the Planck - Einstein relation and can be used to find the energy of a photon that has
You don't need to use this equation in order to find the wavelength of the photon, but you can rewrite it as a function of wavelength
#E = h * c/(lamda)#
This means that the wavelength of the photon can be written in terms of its wavelength as
#lamda = (h * c)/E#