Question

What is the energy of a photon that emits a light of frequency `4.47 times 10^14` `Hz`?

Answer 1

`E=2.96xx10^(-19) J`

Explanation:

`E=hnu`

`E=(6.626xx10^(-34)Js)(4.47xx10^(14) Hz)`

`E= 2.96xx10^(-19) J`

Answer 2

The energy of the photon of light is `2.96xx"10"^(-19)" J"`.

Explanation:

The energy of a photon of a particular wavelength is determined using the following formula:

`E=hnu`,

where:

`E` is energy, `h` is Planck's constant `(6.626xx"10"^(−34)"J"*"s")`, and `nu` is the frequency `(4.47xx"10"^(14)"Hz")` or `((4.47xx10^(14))/("s"))`.

Plug the data into the formula and solve.

`E=6.626xx"10"^(−34)"J"*color(red)cancel(color(black)("s"))xx(4.47xx10^(14))/(color(red)cancel(color(black)("s")))=2.96xx"10"^(-19)" J"` (rounded to three significant figures)

The energy of the photon of light is `2.96xx"10"^(-19)" J"`.