Question

The wavelength of a certain portion of the electromagnetic spectrum ranges from 2000 nm to 3000 nm. The highest frequency associated with the above portion of the spectrum is____________\(\times10^{8}\) MHz (rounded off to one decimal place).

Hint:

Highest frequency comes from the shortest wavelength; use f = c/lambda.

Answer 1

Correct Answer: 1.5

Step 1: Recall the inverse link between frequency and wavelength.
Light of a shorter wavelength always carries a higher frequency, because $f=c/\lambda$ and $c$ stays fixed. So to get the highest frequency in the band $2000$ nm to $3000$ nm, we should use the shortest wavelength, $2000$ nm, not the longest one.

Step 2: Cross check by trying both ends.
At $\lambda=2000$ nm, $f=\dfrac{3\times10^8}{2\times10^{-6}}=1.5\times10^{14}$ Hz.
At $\lambda=3000$ nm, $f=\dfrac{3\times10^8}{3\times10^{-6}}=1.0\times10^{14}$ Hz.
Comparing the two, $1.5\times10^{14}$ Hz is larger, confirming the shorter wavelength gives the higher frequency.

Step 3: Change units from hertz to megahertz.
One megahertz is $10^6$ hertz, so we divide by $10^6$:
\[ 1.5\times10^{14} \div 10^6 = 1.5\times10^{8} \text{ MHz} \]

Step 4: State the final coefficient.
Written in the form asked, the blank multiplying $10^8$ MHz is filled by $1.5$.
$\boxed{1.5\times10^{8} \text{ MHz}}$