Question
The wavelengths of visible light vary from about 3×10−7 m
to 7×10−7 m
. What is the range of frequencies of visible light in a vacuum?(1 point)
Responses
90 Hz
to 210 Hz
90 Hz to 210 Hz
9⋅1010 Hz
to 2.1⋅1011 Hz
9 times 10 to the 10th power Hz to 2 point 1 times 10 to the 11th power Hz
4.28⋅105 Hz
to 1⋅106 Hz
4 point 2 8 times 10 to the 5th power Hz to 1 times 10 to the 6th power Hz
4.29⋅1014 Hz
to 1⋅1015 Hz
to 7×10−7 m
. What is the range of frequencies of visible light in a vacuum?(1 point)
Responses
90 Hz
to 210 Hz
90 Hz to 210 Hz
9⋅1010 Hz
to 2.1⋅1011 Hz
9 times 10 to the 10th power Hz to 2 point 1 times 10 to the 11th power Hz
4.28⋅105 Hz
to 1⋅106 Hz
4 point 2 8 times 10 to the 5th power Hz to 1 times 10 to the 6th power Hz
4.29⋅1014 Hz
to 1⋅1015 Hz
Answers
GPT-4o mini
To find the range of frequencies of visible light in a vacuum, we can use the equation that relates the speed of light, wavelength, and frequency:
\[
c = \lambda \cdot f
\]
where:
- \( c \) is the speed of light, approximately \( 3 \times 10^8 \) m/s,
- \( \lambda \) is the wavelength,
- \( f \) is the frequency.
To find the frequency, we can rearrange the equation to solve for \( f \):
\[
f = \frac{c}{\lambda}
\]
Given the range of wavelengths for visible light:
- Lower bound: \( \lambda = 3 \times 10^{-7} \) m (or 300 nm),
- Upper bound: \( \lambda = 7 \times 10^{-7} \) m (or 700 nm).
### Calculation of Frequencies
1. **For the lower wavelength (3 × 10⁻⁷ m)**:
\[
f_{min} = \frac{3 \times 10^8 \, \text{m/s}}{3 \times 10^{-7} \, \text{m}} = 1 \times 10^{15} \, \text{Hz}
\]
2. **For the upper wavelength (7 × 10⁻⁷ m)**:
\[
f_{max} = \frac{3 \times 10^8 \, \text{m/s}}{7 \times 10^{-7} \, \text{m}} \approx 4.29 \times 10^{14} \, \text{Hz}
\]
### Conclusion
Thus, the frequency range of visible light in a vacuum is approximately from \( 4.29 \times 10^{14} \, \text{Hz} \) to \( 1 \times 10^{15} \, \text{Hz} \).
The correct response is:
**4.29⋅10¹⁴ Hz to 1⋅10¹⁵ Hz**.
\[
c = \lambda \cdot f
\]
where:
- \( c \) is the speed of light, approximately \( 3 \times 10^8 \) m/s,
- \( \lambda \) is the wavelength,
- \( f \) is the frequency.
To find the frequency, we can rearrange the equation to solve for \( f \):
\[
f = \frac{c}{\lambda}
\]
Given the range of wavelengths for visible light:
- Lower bound: \( \lambda = 3 \times 10^{-7} \) m (or 300 nm),
- Upper bound: \( \lambda = 7 \times 10^{-7} \) m (or 700 nm).
### Calculation of Frequencies
1. **For the lower wavelength (3 × 10⁻⁷ m)**:
\[
f_{min} = \frac{3 \times 10^8 \, \text{m/s}}{3 \times 10^{-7} \, \text{m}} = 1 \times 10^{15} \, \text{Hz}
\]
2. **For the upper wavelength (7 × 10⁻⁷ m)**:
\[
f_{max} = \frac{3 \times 10^8 \, \text{m/s}}{7 \times 10^{-7} \, \text{m}} \approx 4.29 \times 10^{14} \, \text{Hz}
\]
### Conclusion
Thus, the frequency range of visible light in a vacuum is approximately from \( 4.29 \times 10^{14} \, \text{Hz} \) to \( 1 \times 10^{15} \, \text{Hz} \).
The correct response is:
**4.29⋅10¹⁴ Hz to 1⋅10¹⁵ Hz**.