Question

Question #b83fb

Answer 1

A) `3.25 xx 10^(-19)J`
B) `5.00 xx 10^14s^(-1)`
C) `612 nm` (Red light)

Explanation:

A) An `eV = 1.6 xx 10^(-19)J` ; `8.82 – 6.79 = 2.03eV` ;
`2.03 xx 1.6 xx 10^(-19)J = 3.25 xx 10^(-19)J`

B) ν = c/λ
Calculate the wavelength required for the specified energy, then calculate the corresponding frequency.

`ν = (3.25 xx 10^(-19))/(6.63 xx 10^(-34) Js)`

`ν = 5.00 xx 10^14s^(-1)`

C) `E_"photon" = (hc)/λ`
where:
`h = 6.63 xx 10^(-34) Js`
`c = 3.0 xx 10^8 m/s`
`λ = m (or nm xx 10^(-9))`

`3.25 xx 10^(-19)J = (6.63 xx 10^(-34) J xx 3.0 xx 10^8)/λ`

`λ = (6.63 xx 10^(-34) J xx 3.0 xx 10^8)/(3.25 xx 10^(-19)J)`

`λ = 6.12 xx 10^(-7)m`

`612 nm` (Red light)