When a radio telescope observes a region of space between two stars it detects electromagnetic radiation that has a wavelength of 0.21 m . This radiation was emitted by hydrogen atoms in the gas and dust located in that region. What is the frequency of this radiation ?

Options

A. `7.1xx10^(10)` Hz
B. `2.1xx10^(14)`Hz
C. `1.4xx10^(9)` Hz
D. `6.9xx10^(11)` Hz

Answer

To find the frequency of the electromagnetic radiation detected by the radio telescope, we can use the relationship between wavelength (λ), frequency (f), and the speed of light (c). The formula is given by:

\[ c = \lambda \cdot f \]

Where:
- \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \) m/s,
- \( \lambda \) is the wavelength,
- \( f \) is the frequency.

### Step-by-Step Solution:

1. **Identify the given values**:
- Wavelength \( \lambda = 0.21 \) m
- Speed of light \( c = 3 \times 10^8 \) m/s

2. **Rearrange the formula to solve for frequency**:
\[ f = \frac{c}{\lambda} \]

3. **Substitute the known values into the equation**:
\[ f = \frac{3 \times 10^8 \text{ m/s}}{0.21 \text{ m}} \]

4. **Calculate the frequency**:
\[ f = \frac{3 \times 10^8}{0.21} \]
\[ f \approx 1.42857 \times 10^9 \text{ Hz} \]

5. **Round the frequency to two decimal places**:
\[ f \approx 1.43 \times 10^9 \text{ Hz} \]

### Final Answer:
The frequency of the radiation is approximately \( 1.43 \times 10^9 \) Hz.

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