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Question 9 of 10

What frequency corresponds to an absorption line at [tex][tex]$527 \text{ nm}$[/tex][/tex]?

A. [tex][tex]$1.76 \times 10^{14} \text{ Hz}$[/tex][/tex]
B. [tex][tex]$5.69 \times 10^{14} \text{ Hz}$[/tex][/tex]
C. [tex][tex]$6.71 \times 10^{14} \text{ Hz}$[/tex][/tex]
D. [tex][tex]$3.77 \times 10^{14} \text{ Hz}$[/tex][/tex]

Sagot :

To determine the frequency that corresponds to an absorption line at [tex]\( 527 \, \text{nm} \)[/tex], follow these steps:

1. Understand the relationship between wavelength and frequency:
- The speed of light ([tex]\( c \)[/tex]) is approximately [tex]\( 3 \times 10^8 \, \text{m/s} \)[/tex].
- The formula to relate wavelength ([tex]\( \lambda \)[/tex]) and frequency ([tex]\( f \)[/tex]) is:
[tex]\[ f = \frac{c}{\lambda} \][/tex]

2. Convert the wavelength from nanometers to meters:
- The given wavelength is [tex]\( 527 \, \text{nm} \)[/tex]. Since [tex]\( 1 \, \text{nm} = 10^{-9} \, \text{m} \)[/tex]:
[tex]\[ 527 \, \text{nm} = 527 \times 10^{-9} \, \text{m} \][/tex]

3. Calculate the frequency:
- Using the speed of light ([tex]\( c \)[/tex]) and the converted wavelength ([tex]\( \lambda \)[/tex]), the frequency ([tex]\( f \)[/tex]) is calculated as:
[tex]\[ f = \frac{3 \times 10^8 \, \text{m/s}}{527 \times 10^{-9} \, \text{m}} \][/tex]

4. Compare the calculated frequency with the given choices:
- The calculated frequency should be compared with the following given choices:
- [tex]\(\text{A. } 1.76 \times 10^{14} \, \text{Hz}\)[/tex]
- [tex]\(\text{B. } 5.69 \times 10^{14} \, \text{Hz}\)[/tex]
- [tex]\(\text{C. } 6.71 \times 10^{14} \, \text{Hz}\)[/tex]
- [tex]\(\text{D. } 3.77 \times 10^{14} \, \text{Hz}\)[/tex]

After going through these steps, we find that none of the given choices match the calculated frequency closely. Therefore, the correct answer to this question is:

[tex]\[ 0 \quad \text{(indicating no match found)} \][/tex]

Thus, the frequency that corresponds to an absorption line at [tex]\( 527 \, \text{nm} \)[/tex] does not match any of the given choices.