Class 10 - Unit # 12 : Electromagnetic Spectrum - Solved Numericals

Electromagnetic Waves

1. Electromagnetic radiation having a $15.0\text{ }\mu\text{m}$ wavelength is classified as infrared radiation. What is its frequency? Given that the speed of light is $3 \times 10^8\text{ m/s}$.

Data:

• Wavelength ($\lambda$) = $15.0\text{ }\mu\text{m} = 15.0 \times 10^{-6}\text{ m}$
• Speed of light ($c$) = $3 \times 10^8\text{ m/s}$
• Frequency ($f$) = ?

Solution:

According to the wave equation for electromagnetic radiation:

$$c = f \cdot \lambda$$ $$f = \frac{c}{\lambda}$$ $$f = \frac{3 \times 10^8\text{ m/s}}{15.0 \times 10^{-6}\text{ m}}$$ $$f = 0.2 \times 10^{14}\text{ Hz} = 2.0 \times 10^{13}\text{ Hz}$$

RESULT: The frequency of the infrared radiation is $2.0 \times 10^{13}\text{ Hz}$.

2. What is the frequency of the $193\text{ nm}$ ultraviolet radiation used in laser eye surgery?

Data:

• Wavelength ($\lambda$) = $193\text{ nm} = 193 \times 10^{-9}\text{ m}$
• Speed of light ($c$) = $3 \times 10^8\text{ m/s}$
• Frequency ($f$) = ?

Solution:

Using the electromagnetic wave equation:

$$c = f \cdot \lambda$$ $$f = \frac{c}{\lambda}$$ $$f = \frac{3 \times 10^8\text{ m/s}}{193 \times 10^{-9}\text{ m}}$$ $$f \approx 0.01554 \times 10^{17}\text{ Hz} = 1.55 \times 10^{15}\text{ Hz}$$

RESULT: The frequency of the ultraviolet radiation is $1.55 \times 10^{15}\text{ Hz}$.

3. Calculate the wavelength of $100\text{-MHz}$ radio waves used in an MRI unit?

Data:

• Frequency ($f$) = $100\text{ MHz} = 100 \times 10^6\text{ Hz} = 10^8\text{ Hz}$
• Speed of light ($c$) = $3 \times 10^8\text{ m/s}$
• Wavelength ($\lambda$) = ?

Solution:

Rearranging the wave equation to isolate wavelength:

$$c = f \cdot \lambda$$ $$\lambda = \frac{c}{f}$$ $$\lambda = \frac{3 \times 10^8\text{ m/s}}{10^8\text{ Hz}}$$ $$\lambda = 3\text{ m}$$

RESULT: The wavelength of the radio waves used is $3\text{ meters}$.

6. What is the frequency of green light with a wavelength of $5.5 \times 10^{-7}\text{ m}$?

Data:

• Wavelength ($\lambda$) = $5.5 \times 10^{-7}\text{ m}$
• Speed of light ($c$) = $3 \times 10^8\text{ m/s}$
• Frequency ($f$) = ?

Solution:

Using the wave property formula:

$$c = f \cdot \lambda$$ $$f = \frac{c}{\lambda}$$ $$f = \frac{3 \times 10^8\text{ m/s}}{5.5 \times 10^{-7}\text{ m}}$$ $$f \approx 5.45 \times 10^{14}\text{ Hz}$$

RESULT: The frequency of the radiation is $5.45 \times 10^{14}\text{ Hz}$.

7. A typical household microwave oven operates at a frequency of $2.45\text{-GHz}$. What is the wavelength of this radiation?

Data:

• Frequency ($f$) = $2.45\text{ GHz} = 2.45 \times 10^9\text{ Hz}$
• Speed of light ($c$) = $3 \times 10^8\text{ m/s}$
• Wavelength ($\lambda$) = ?

Solution:

Solving for wavelength:

$$c = f \cdot \lambda$$ $$\lambda = \frac{c}{f}$$ $$\lambda = \frac{3 \times 10^8\text{ m/s}}{2.45 \times 10^9\text{ Hz}}$$ $$\lambda \approx 0.122\text{ m}\text{ (or } 12.2\text{ cm)}$$

RESULT: The wavelength of the microwaves used is $0.122\text{ meters}$.

Space & Distance Dynamics

4. The distance from Earth to the Sun is $1.49 \times 10^{11}\text{ meters}$. How long does a radio pulse radiated from the Sun take to reach the Earth?

Data:

• Distance ($s$) = $1.49 \times 10^{11}\text{ m}$
• Velocity of wave ($v = c$) = $3 \times 10^8\text{ m/s}$
• Time taken ($t$) = ?

Solution:

According to the standard equation for uniform rectilinear motion:

$$s = v \cdot t$$ $$t = \frac{s}{c}$$ $$t = \frac{1.49 \times 10^{11}\text{ m}}{3 \times 10^8\text{ m/s}}$$ $$t \approx 496.67\text{ seconds}$$

Optional conversion to minutes: $\frac{496.67}{60} \approx 8\text{ minutes and } 17\text{ seconds}$.

RESULT: The time taken by the radiation is $496.7\text{ seconds}$.

5. Distances in space are often measured in units of light-years, the distance light travels in one year. Find the distance in kilometers in a light-year.

Data:

• Speed of light ($c$) = $3 \times 10^8\text{ m/s}$
• Time ($t$) = $1\text{ year} = 365 \times 24 \times 60 \times 60\text{ s} = 31,536,000\text{ s}$
• Distance ($s$) = ?

Solution:

First, calculate the total distance traveled in meters:

$$s = c \cdot t$$ $$s = (3 \times 10^8\text{ m/s}) \times (31,536,000\text{ s})$$ $$s = 9.46 \times 10^{15}\text{ meters}$$

Now, convert the value into kilometers (since $1\text{ km} = 10^3\text{ m}$):

$$s_{\text{km}} = \frac{9.46 \times 10^{15}\text{ m}}{1000}$$ $$s_{\text{km}} = 9.46 \times 10^{12}\text{ km}$$

RESULT: One light-year is a distance equal to $9.46 \times 10^{12}\text{ km}$ (approx. $9.5\text{ trillion kilometers}$).