Question

Excited lithium ions emit radiation at a wavelength of 670.8 nm in the visible range spectrum. Calculate the frequency of a photon of this radiation

Answer 1

`f=4.47x10^5GHz`

Step-by-step explanation:

Hello,

In this case, we relate the speed of light, wavelength and frequency via the shown below equation expressed in the proper SI system of units:

`f=(c)/(\lambda ) =(3x10^8m/s)/(670.8nm*(1x10^(-9)m)/(1nm) ) =4.47x10^(14)Hz*(1GHz)/(1x10^9Hz)\\ f=4.47x10^5GHz`

Best regards.

Answer 2

Answer : The frequency of a photon of radiation is,
`4.47* 10^9s^(-1)`

Explanation : Given,

Wavelength of the radiation = 670.8 nm

First we have to convert wavelength form 'nm' to 'm'.

Conversion used :
`(1nm=10^(-9)m)`

So, the wavelength of the radiation = 670.8 nm =
`670.8* 10^(-9)m`

Now we have to calculate the frequency of a photon of radiation.

Formula used :
`\\u =(c)/(\lambda)`

where,

`\\u` = frequency of a photon of radiation

`\lambda` = wavelength of the radiation

c = speed of light =
`3* 10^8m/s`

Now put all the given values in the above formula, we get the frequency of a photon of radiation.

`\\u =(3* 10^8m/s)/(670.8* 10^(-9)m)`

`\\u =4.47* 10^9s^(-1)`

Therefore, the frequency of a photon of radiation is,
`4.47* 10^9s^(-1)`