To find the wavelength of a photon given its energy, we will use the relationship between energy and wavelength in the context of quantum mechanics. The key formula here is derived from the energy of a photon:
[tex]\[ E = \frac{hc}{\lambda} \][/tex]
Where:
- [tex]\( E \)[/tex] is the energy of the photon.
- [tex]\( h \)[/tex] is Planck's constant.
- [tex]\( c \)[/tex] is the speed of light.
- [tex]\( \lambda \)[/tex] is the wavelength of the photon.
Given values:
- [tex]\( E = 2 \)[/tex] electron volts (eV)
Constants:
- Planck's constant, [tex]\( h = 4.135667696 \times 10^{-15} \)[/tex] eV·s
- Speed of light, [tex]\( c = 299,792,458 \)[/tex] m/s
Step-by-Step Solution:
1. Rearrange the formula to solve for wavelength ([tex]\( \lambda \)[/tex]):
[tex]\[ \lambda = \frac{hc}{E} \][/tex]
2. Substitute the known values into the formula:
Plug in [tex]\( h = 4.135667696 \times 10^{-15} \)[/tex] eV·s, [tex]\( c = 299,792,458 \)[/tex] m/s, and [tex]\( E = 2 \)[/tex] eV.
[tex]\[ \lambda = \frac{(4.135667696 \times 10^{-15} \, \text{eV} \cdot \text{s}) \times (299,792,458 \, \text{m/s})}{2 \, \text{eV}} \][/tex]
3. Perform the multiplication and division:
[tex]\[ \lambda = \frac{4.135667696 \times 299,792,458}{2} \times 10^{-15} \, \text{m} \][/tex]
[tex]\[ \lambda = \frac{1,239,841,984}{2} \times 10^{-15} \, \text{m} \][/tex]
[tex]\[ \lambda = 619.9209920275184 \times 10^{-9} \, \text{m} \][/tex]
4. Convert the wavelength from meters to nanometers (1 nm = [tex]\( 10^{-9} \)[/tex] meters):
[tex]\[ \lambda = 619.9209920275184 \, \text{nm} \][/tex]
Thus, the wavelength of a photon with an energy of 2 eV is approximately:
[tex]\[ 619.92 \, \text{nm} \][/tex]
