To determine the frequency and energy of light given the wavelength, we can use two fundamental equations from physics. The first is the relationship between the speed of light (c), the frequency (ν), and the wavelength (λ):
[tex]\[ c = \nu \lambda \][/tex]
The second is the relationship between the energy (E) of a photon and its frequency:
[tex]\[ E = h\nu \][/tex]
where:
- [tex]\( c \)[/tex] is the speed of light in a vacuum ([tex]\( 3.0 \times 10^8 \)[/tex] m/s),
- [tex]\( h \)[/tex] is Planck's constant ([tex]\( 6.626 \times 10^{-34} \)[/tex] J·s).
Given:
- Wavelength of lithium (Li): [tex]\( 671 \)[/tex] nm.
Steps:
### Step 1: Convert the wavelength from nanometers to meters
[tex]\[ 671 \, \text{nm} = 671 \times 10^{-9} \, \text{m} = 6.71 \times 10^{-7} \, \text{m} \][/tex]
### Step 2: Calculate the frequency
Using the equation [tex]\( \nu = \frac{c}{\lambda} \)[/tex]:
[tex]\[ \nu = \frac{3.0 \times 10^8 \, \text{m/s}}{6.71 \times 10^{-7} \, \text{m}} \][/tex]
[tex]\[ \nu \approx 4.47 \times 10^{14} \, \text{Hz} \][/tex]
### Step 3: Calculate the energy
Using the equation [tex]\( E = h \nu \)[/tex]:
[tex]\[ E = (6.626 \times 10^{-34} \, \text{J·s}) \times (4.47 \times 10^{14} \, \text{Hz}) \][/tex]
[tex]\[ E \approx 2.96 \times 10^{-19} \, \text{J} \][/tex]
### Summary
For the given wavelength of lithium:
- Wavelength ([tex]\(\lambda\)[/tex]): [tex]\( 6.71 \times 10^{-7} \, \text{m} \)[/tex]
- Frequency ([tex]\(\nu\)[/tex]): [tex]\( 4.47 \times 10^{14} \, \text{Hz} \)[/tex]
- Energy ([tex]\(E\)[/tex]): [tex]\( 2.96 \times 10^{-19} \, \text{J} \)[/tex]
These values suggest that the light emitted by lithium will be in the visible region of the electromagnetic spectrum, specifically in the red part, as wavelengths around 671 nm correspond to red light.
