Q.A radio can tune in to any station in the $7.5\ \text{MHz}$ to $12\ \text{MHz}$ band. What is the corresponding wavelength band?

The key idea is the wave equation: for any electromagnetic wave, $c = f \lambda$, where $c = 3 \times 10^8\ \text{m/s}$ is the speed of light.

Step 1: For the lowest frequency $f_1 = 7.5\ \text{MHz} = 7.5 \times 10^6\ \text{Hz}$, the wavelength is

Step 2: For the highest frequency $f_2 = 12\ \text{MHz} = 12 \times 10^6\ \text{Hz}$, the wavelength is

$\lambda_2 = \frac{c}{f_2} = \frac{3 \times 10^8}{12 \times 10^6} = 25\ \text{m}.$ …

The key idea here is beautifully simple: all electromagnetic waves — including radio waves — travel at the same speed in vacuum (and very nearly the same in air). That speed is $c = 3.0 \times 10^8\ \text{m/s}$. The relationship between frequency $f$ and wavelength $\lambda$ is:

So if you know the frequency, you get the wavelength by $\lambda = c/f$. And because $c$ is constant, a higher frequency means a shorter wavelength, and vice versa. That inverse relationship is the heart of this problem.

Now, the radio tunes from $7.5\ \text{MHz}$ to $12\ \text{MHz}$. "MHz" means megahertz, or $10^6\ \text{Hz}$. So:

• Lower frequency: $f_1 = 7.5 \times 10^6\ \text{Hz}$
• Upper frequency: $f_2 = 12 \times 10^6\ \text{Hz}$

We want the corresponding wavelength band. Since wavelength is inversely proportional to frequency, the longest wavelength comes from the lowest frequency, and the shortest wavelength from the highest frequency.

1. Longest wavelength (at $f_1 = 7.5\ \text{MHz}$):

Divide the numbers: $3.0 / 7.5 = 0.4$, and $10^8 / 10^6 = 10^2$. So:

2. Shortest wavelength (at $f_2 = 12\ \text{MHz}$):

$3.0 / 12 = 0.25$, and again $10^8 / 10^6 = 10^2$. So:

$\lambda_{\text{min}} = 0.25 \times 10^2 = 25\ \text{m}$ …

Method: Wave Equation Method

This problem uses the fundamental relationship between wave speed, frequency, and wavelength.

Concept First

For any electromagnetic wave (including radio waves), the speed is constant in vacuum/air:

• Speed of light: $c = 3 \times 10^8\ \text{m/s}$
• Relation: $c = f \lambda$, where $f$ = frequency, $\lambda$ = wavelength

Steps

Step 1: Write the wave equation

Step 2: Convert frequencies to Hz

• Lower frequency: $7.5\ \text{MHz} = 7.5 \times 10^6\ \text{Hz}$
• Upper frequency: $12\ \text{MHz} = 12 \times 10^6\ \text{Hz}$

Step 3: Calculate wavelengths

For lower frequency (longer wavelength):

$\lambda_1 = \frac{3 \times 10^8}{7.5 \times 10^6} = \mathbf{40\ \text{m}}$ …

Here are the most common mistakes students make when solving this type of problem, along with how to avoid each.

Mistake 1: Forgetting the relationship between frequency and wavelength

The error: Students often try to guess or recall a formula incorrectly, mixing up $c = f\lambda$ with other equations (like $v = f\lambda$ for sound, or $E = hf$). They might also forget that frequency and wavelength are inversely proportional for a wave traveling at constant speed.

How to avoid:

• Always start with the fundamental relation:

where $c = 3 \times 10^8\ \text{m/s}$ (speed of light in vacuum/air).

• Understand the concept: For a fixed speed, if frequency increases, wavelength decreases. This helps you check if your answer makes sense — the higher frequency end of the band should give the smaller wavelength.

Mistake 2: Using the wrong units or forgetting to convert

The error: Students plug in $7.5\ \text{MHz}$ directly as $7.5$ without converting to Hz. This gives a wildly wrong wavelength (off by a factor of $10^6$).

How to avoid:

• Always convert MHz to Hz:

So $7.5\ \text{MHz} = 7.5 \times 10^6\ \text{Hz}$ and $12\ \text{MHz} = 12 \times 10^6\ \text{Hz}$.

• Write the conversion step explicitly in your solution — don’t do it mentally.

Mistake 3: Calculating only one wavelength (instead of the band)

The error: Students compute the wavelength for only one frequency (e.g., $7.5\ \text{MHz}$) and assume that’s the answer. The question asks for the corresponding wavelength band, which is a range.

How to avoid:

• Identify that you need two calculations:
  • For the lower frequency ($7.5\ \text{MHz}$), you get the longer wavelength.
  • For the higher frequency ($12\ \text{MHz}$), you get the shorter wavelength.
• State the answer as a range:

Mistake 4: Reversing the order of the wavelength band

The error: Students write the band as $25\ \text{m}$ to $40\ \text{m}$ instead of $40\ \text{m}$ to $25\ \text{m}$ (or vice versa). Since wavelength decreases as frequency increases, the band should be written from longest wavelength to shortest wavelength (or clearly labeled).

How to avoid:

• Check the logic: Lower frequency → longer wavelength. So the $7.5\ \text{MHz}$ end gives the larger wavelength.
• Write the answer clearly: …