Question

Question #a3629

Answer 1

`E=2.29xx10^-19J`

Explanation:

Energy of photon = Planck's constant x frequency or `E=hf`.

Because `f` is not given, we use a different equation to find `f`. That is wave equation, which is `c=flambda`

where,

`c` is the speed of light/photon`= 3.00 xx 10^8 m.s`
`lambda` is the wavelength `(m)`
`f` is the frequency`(H_z)`

When we rearrange this formula to solve for `f`, we get `f=c/lambda`.

Keep in mind that wavelength is measured in metres, therefore we need to convert #=869nm to metres.

Nano is `10^-9`,
`869 xx 10^-9=8.69xx10^-7m`

`f=(3.00 xx 10^8)/(8.69 xx 10^-7)`
`= 3.45xx10^14H_z`

Now we are able to find `E.`

`=(6.63 xx 10^-34) xx ( 3.45 xx 10^14)`

`E=2.29xx10^-19J`