Question

Question #80df2

Answer 1

Here's what I got.

Explanation:

For starters, you should know that wavelength and frequency have an inverse relationship that can be described using 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)`

This means that photons that have a higher frequency will have a shorter wavelength and photons that have a lower frequency will have a longer wavelength.

Notice that the speed of light is expressed in meters per second, so make sure to convert the wavelength from nanometers to meters before using the equation.

`710 color(red)(cancel(color(black)("nm"))) * "1 m"/(10^9 color(red)(cancel(color(black)("nm")))) = 7.10 * 10^(-7)` `"m"`

Rearrange the equation to solve for `nu`, the frequency of the photon

`lamda * nu = 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 * 10^(-7)color(red)(cancel(color(black)("m")))) = color(darkgreen)(ul(color(black)(4.2 * 10^(14) color(white)(.)"s"^(-1))))`

Now, the energy of the photon is directly proportional to its frequency, 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"`
• `nu` is the frequency of the photon

This time, you can say that photons that have a higher frequency will have a higher energy than photons that have a lower frequency.

Plug in your value to find

`E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 4.2 * 10^(14)color(red)(cancel(color(black)("s"^(-1))))`

`E = color(darkgreen)(ul(color(black)(2.8 * 10^(-19)color(white)(.)"J")))`

Both values are rounded to two sig figs, the number of significant figures you have for the wavelength of the photon.