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: Identify which wavelength gives the highest frequency.
Frequency and wavelength are inversely related through the wave equation
\[ f = \frac{c}{\lambda} \]
where \(c\) is the speed of light and \(\lambda\) is the wavelength. Since frequency goes up as wavelength goes down, the highest frequency in the given range corresponds to the shortest wavelength, which is \(2000\) nm.

Step 2: Convert the wavelength to metres.
\[ \lambda = 2000 \text{ nm} = 2000 \times 10^{-9} \text{ m} = 2 \times 10^{-6} \text{ m} \]

Step 3: Apply the wave equation with \(c = 3\times10^{8}\) m/s.
\[ f = \frac{c}{\lambda} = \frac{3\times10^{8}}{2\times10^{-6}} = 1.5\times10^{14} \text{ Hz} \]

Step 4: Convert the frequency from Hz to MHz.
Since \(1\) MHz \(= 10^{6}\) Hz,
\[ f = \frac{1.5\times10^{14}}{10^{6}} \text{ MHz} = 1.5\times10^{8} \text{ MHz} \]

Step 5: Final conclusion.
So the highest frequency in the given range of the electromagnetic spectrum is
\[ \boxed{1.5\times10^{8} \text{ MHz}} \]
The required coefficient, rounded to one decimal place, is \(1.5\).