Question

Question #57b52

Answer 1

For part (a) `E = 4.0 * 10^(-21)"J"`

Explanation:

I'll show you how to calculate the energy of the photon given to you in part (a), and leave the other one to you as practice.

So, what do you know about the energy of a photon?

You know that its proportional to the frequency of the photon, `nu`.

`E = nu *h "`, where

`h` - Planck's constant, equal to `6.626 * 10^(-34)"J" * "s"`

Since you are given the wavelength of the photon, you can calculate its frequency by using the inverse relationship that exists between wavelength, `lamda`, and frequency, `nu`

`lamda * nu = c implies nu = c/(lamda)" "`, where

`c` - the speed of light in a vacuum, approximately equal to `3.0 * 10^(8)"ms"^(-1)`

Use this relationship in the equation for the energy of the photon to get

`E = c/(lamda) * h`

Plug in your values to find `E` - do not forget to convert the wavelength of the photon from nanometers to meters!

`E= (3.0 * 10^(8)color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))) * 6.626 * 10^(-34)"J"color(red)(cancel(color(black)("s"))))/(5.0 * 10^4 * 10^(-9)color(red)(cancel(color(black)("m"))))`

`E = 3.976 * 10^(-21)"J"`

Round this off to two sig figs, the number of sig figs you have for the wavelength of the photon, to get

`E = color(green)(4.0 * 10^(-21)"J")`