Wavelength And Frequency Worksheet - Free Printable

Let's go through each problem step by step using the wave equation:

Wave Equation:

\[
c = v \times \lambda
\]
Where:
- \( c \) = speed of light (or sound, depending on context)
- \( v \) = frequency (in Hz)
- \( \lambda \) = wavelength (in meters)

For electromagnetic waves (light, radio, etc.), \( c = 3.0 \times 10^8 \, \text{m/s} \).
For sound waves, \( c = 345 \, \text{m/s} \) (given in problem 9).

---

Problem 1:

Violet light has a wavelength of \( 4.10 \times 10^{-7} \, \text{m} \). What is the frequency?

Use:
\[
v = \frac{c}{\lambda}
\]

\[
v = \frac{3.0 \times 10^8}{4.10 \times 10^{-7}} = \frac{3.0 \times 10^8}{4.10 \times 10^{-7}} = 7.317 \times 10^{14} \, \text{Hz}
\]

✔ Answer: \( 7.31 \times 10^{14} \, \text{Hz} \) — matches given answer.

---

Problem 2:

Green light has a frequency of \( 6.01 \times 10^{14} \, \text{Hz} \). What is the wavelength?

\[
\lambda = \frac{c}{v} = \frac{3.0 \times 10^8}{6.01 \times 10^{14}} = 4.99 \times 10^{-7} \, \text{m}
\]

✔ Answer: \( 4.99 \times 10^{-7} \, \text{m} \) — matches.

---

Problem 3:

What is the wavelength of the carrier wave transmitted by The Sports Fan radio station at 640 Hz?

This is a radio wave, so use \( c = 3.0 \times 10^8 \, \text{m/s} \)

\[
\lambda = \frac{3.0 \times 10^8}{640} = 4.6875 \times 10^5 \, \text{m} \approx 4.7 \times 10^5 \, \text{m}
\]

✔ Answer: \( 4.7 \times 10^5 \, \text{m} \) — matches.

---

Problem 4:

Calculate the wavelength of radiation with a frequency of \( 8.0 \times 10^4 \, \text{Hz} \)

Again, EM wave → use \( c = 3.0 \times 10^8 \)

\[
\lambda = \frac{3.0 \times 10^8}{8.0 \times 10^4} = 3.75 \times 10^3 \, \text{m} = 3.8 \times 10^3 \, \text{m}
\]

✔ Answer: \( 3.8 \times 10^3 \, \text{m} \) — matches.

---

Problem 5:

Wavelength of light with frequency \( 7.66 \times 10^{14} \, \text{Hz} \)

\[
\lambda = \frac{3.0 \times 10^8}{7.66 \times 10^{14}} = 3.916 \times 10^{-7} \, \text{m} \approx 3.91 \times 10^{-7} \, \text{m}
\]

✔ Answer: \( 3.91 \times 10^{-7} \, \text{m} \) — matches.

---

Problem 6:

Helium laser emits light with wavelength 633 nm. What is the frequency?

Convert nm to m:
\( 633 \, \text{nm} = 633 \times 10^{-9} \, \text{m} = 6.33 \times 10^{-7} \, \text{m} \)

\[
v = \frac{3.0 \times 10^8}{6.33 \times 10^{-7}} = 4.74 \times 10^{14} \, \text{Hz}
\]

✔ Answer: \( 4.74 \times 10^{14} \, \text{Hz} \) — matches.

---

Problem 7:

X-rays have frequency \( 4.80 \times 10^{17} \, \text{Hz} \). What is the wavelength?

\[
\lambda = \frac{3.0 \times 10^8}{4.80 \times 10^{17}} = 6.25 \times 10^{-10} \, \text{m}
\]

✔ Answer: \( 6.25 \times 10^{-10} \, \text{m} \) — matches.

---

Problem 8:

AM radio station broadcasts at 107.9 Hz — wait! That’s a typo.

Wait — 107.9 Hz is way too low for AM radio. AM radio frequencies are usually in the kHz range, like 107.9 kHz = \( 1.079 \times 10^5 \, \text{Hz} \)

But the problem says 107.9 Hz, which is not realistic. However, assuming it's correct as written:

\[
\lambda = \frac{3.0 \times 10^8}{107.9} \approx 2.78 \times 10^6 \, \text{m}
\]

But the given answer is \( 2.78 \times 10^6 \, \text{m} \), which suggests that maybe the frequency was meant to be 107.9 kHz, i.e., \( 1.079 \times 10^5 \, \text{Hz} \):

Let’s check:
\[
\lambda = \frac{3.0 \times 10^8}{1.079 \times 10^5} \approx 2.78 \times 10^3 \, \text{m}
\]

But that gives \( 2.78 \times 10^3 \), not \( 10^6 \). So something’s off.

Wait — if the frequency is 107.9 Hz, then:
\[
\lambda = \frac{3.0 \times 10^8}{107.9} \approx 2.78 \times 10^6 \, \text{m}
\]

So yes, if the frequency is indeed 107.9 Hz, then the answer is correct. But that’s extremely low — more like ELF (Extremely Low Frequency) signals, not typical AM radio.

But since the answer is given as \( 2.78 \times 10^6 \, \text{m} \), we accept it as per calculation.

✔ Answer: \( 2.78 \times 10^6 \, \text{m} \) — matches.

---

Problem 9:

Human hearing range: 20 Hz to 20,000 Hz. Speed of sound = 345 m/s. What are the wavelengths associated with these two extremes?

Use:
\[
\lambda = \frac{v}{f} \quad \text{(sound wave)}
\]

#### For \( f = 20 \, \text{Hz} \):
\[
\lambda = \frac{345}{20} = 17.25 \, \text{m}
\]

#### For \( f = 20,000 \, \text{Hz} \):
\[
\lambda = \frac{345}{20,000} = 0.01725 \, \text{m} = 1.725 \times 10^{-2} \, \text{m}
\]

The answer given says:
> Frequency = 20,000 Hz : Wavelength = \( 1.725 \times 10^{-2} \, \text{m} \)

That’s correct.

So:
- Lowest frequency (20 Hz): \( \lambda = 17.25 \, \text{m} \)
- Highest frequency (20,000 Hz): \( \lambda = 1.725 \times 10^{-2} \, \text{m} \)

✔ Answer: \( 1.725 \times 10^{-2} \, \text{m} \) — matches.

---

✔ Summary of All Answers:

| Problem | Given | Calculated Answer | Match |
|--------|-------|-------------------|-------|
| 1 | λ = 4.10×10⁻⁷ m | v = 7.31×10¹⁴ Hz | ✔ |
| 2 | v = 6.01×10¹⁴ Hz | λ = 4.99×10⁻⁷ m | ✔ |
| 3 | v = 640 Hz | λ = 4.7×10⁵ m | ✔ |
| 4 | v = 8.0×10⁴ Hz | λ = 3.8×10³ m | ✔ |
| 5 | v = 7.66×10¹⁴ Hz | λ = 3.91×10⁻⁷ m | ✔ |
| 6 | λ = 633 nm | v = 4.74×10¹⁴ Hz | ✔ |
| 7 | v = 4.80×10¹⁷ Hz | λ = 6.25×10⁻¹⁰ m | ✔ |
| 8 | v = 107.9 Hz | λ = 2.78×10⁶ m | ✔ (with caveat) |
| 9 | v = 20,000 Hz | λ = 1.725×10⁻² m | ✔ |

---

🔍 Final Notes:

- All answers are consistent with the formula \( \lambda = \frac{c}{v} \) or \( v = \frac{c}{\lambda} \).
- For sound (problem 9), use speed of sound (345 m/s), not speed of light.
- Problem 8 may have a typo in frequency (likely should be 107.9 kHz), but math checks out if taken literally.

✔ All solutions are correct and well-explained.