Question

Question #1c187

Answer 1

The wavelength of the unknown wave is shorter than the wavelength of red light.

Explanation:

Your strategy here will be to use the fact that frequency and wavelength are inversely proportional to calculate the wavelength of the unknown wave

`color(blue)(|bar(ul(color(white)(a/a)nu * lamda = c color(white)(a/a)|)))`

Here

`nu` - the frequency of the wave
`lamda` - its wavelength
`c` - the speed of light in a vacuum , in your case geven as `2.998 * 10^8"m s"^(-1)`

Now, the frequency of the unknown wave is given in hertz, `"Hz"`. A hertz is defined as one cycle per second.

A frequency of

`nu = 5.3 * 10^15"Hz"`

tells you that your unknown wave goes through `5.3 * 10^15` complete cycles per second. You can thus say that the frequency is equal to

`nu = 5.3 * 10^15"s"^(-1)`

So, use the above equation to calculate the wavelength of the unknown wave

`nu * lamda = c implies lamda = c/(nu)`

Plug in your values to find

`lamda = (2.998 * 10^8"m" color(red)(cancel(color(black)("s"^(-1)))))/(5.3 * 10^(15)color(red)(cancel(color(black)("s"^(-1))))) = 5.7 * 10^(-8)"m"`

As you can see, the wavelength of the unknown wave is shorter than the wavelength of red light, since

`color(green)(|bar(ul(color(white)(a/a)color(black)(5.7 * 10^(-8)"m" " "<" " 6.5 * 10^(-7)"m")color(white)(a/a)|)))`

Since wavelength and frequency are inversely proportional, you can say that the frequency of the unknown wave is higher than the frequency of red light.

This places your unknown wave in the ultra-violet portion of the EM spectrum.