Question

Question #39e52

Answer 1

`2.633 x 10^(-19)` Joules - A nice, deep red!

Explanation:

`E_("photon") = hc/lambda`

where: `h = 6.63 x 10^(-34) J*s`
`c = 3.0 x 10^8 m/s`
`lambda = m (or nm x 10^(-9))`

These terms can be combined into a single factor so that all you need to do is put in the wavelength. It is very important to remember that the equation uses a wavelength in METERS, so using the common nanometer wavelength description requires an exponent adjustment.
`E_("photon") = 1.988 x 10^(-16)/(nm)`

Answer 2

`Answer: 2.63 * 10^-19 J`

Explanation:

The energy of a photon is given by the following equation.

`color(white)(aaaaaaaaaaaaa)`Equation(a) `E=h*f`

Where
`h = "Planck's constant" (6.62 * 10^-34 J*s)`
`f = "frequency" (1/s)`

Since we know that the speed of light is constant and formulated by

`c = lambda*f`

Where
`c = 3.00*10^8m/s`
`lambda = "wavelength" (m)`

If we rearrange the equation to find frequency, we can plug this back into Equation (a) to solve for photon Energy.

`c/lambda = f`

`color(white)(aaaaaaaaaaa)`Equation(b) `E = h*c/lambda`

Plug in and solve

`E = ((3.00*10^8 cancelm/cancels)*(6.62*10^-34J*cancels))/(7.55 *10^-7cancelm) = 2.63 * 10^-19`

`Answer: 2.63 * 10^-19 J`