Question

For IR radiation with û = 1,130 cm 1, v=__THz

Answer 1

Answer: The frequency of the radiation is 33.9 THz

Step-by-step explanation:

We are given:

Wave number of the radiation,
`\bar{\\u}=1130cm^(-1)`

Wave number is defined as the number of wavelengths per unit length.

Mathematically,

`\bar{\\u}=(1)/(\lambda)`

where,

`\bar{\\u}` = wave number =
`1130cm^(-1)`

`\lambda` = wavelength of the radiation = ?

Putting values in above equation, we get:

`1130cm^(-1)=\farc{1}{\lambda}\\\\\lambda=(1)/(1130cm^(-1))=8.850* 10^(-4)cm`

Converting this into meters, we use the conversion factor:

1 m = 100 cm

So,
`8.850* 10^(-4)cm=8.850* 10^(-4)* 10^(-2)=8.850* 10^(-6)m`

• The relation between frequency and wavelength is given as:

`\\u=(c)/(\lambda)`

where,

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

`\\u` = frequency of the radiation = ?

Putting values in above equation, we get:

`\\u=(3* 10^8m/s)/(8.850* 10^(-4)m)`

`\\u=0.339* 10^(14)Hz`

Converting this into tera Hertz, we use the conversion factor:

`1THz=1* 10^(12)Hz`

So,
`0.339* 10^(14)Hz* (1THz)/(1* 10^(12)Hz)=33.9THz`

Hence, the frequency of the radiation is 33.9 THz