Alright, let's determine the kinetic energy of electrons given a wavelength of [tex]\(3.45 \times 10^{-3} \, \text{m}\)[/tex].
To find the kinetic energy of the electrons, we will first calculate their energy using the photon energy formula derived from Planck's equation:
[tex]\[ E = \frac{hc}{\lambda} \][/tex]
where:
- [tex]\( E \)[/tex] is the energy,
- [tex]\( h \)[/tex] is Planck’s constant, [tex]\( 6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s} \)[/tex],
- [tex]\( c \)[/tex] is the speed of light, [tex]\( 3.00 \times 10^8 \, \text{m/s} \)[/tex],
- [tex]\( \lambda \)[/tex] is the wavelength, [tex]\( 3.45 \times 10^{-3} \, \text{m} \)[/tex].
Let’s substitute these values into the formula and calculate the energy:
[tex]\[ E = \frac{(6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (3.00 \times 10^8 \, \text{m/s})}{3.45 \times 10^{-3} \, \text{m}} \][/tex]
First, we multiply Planck’s constant [tex]\( h \)[/tex] by the speed of light [tex]\( c \)[/tex]:
[tex]\[ 6.62607015 \times 10^{-34} \times 3.00 \times 10^8 = 1.987821045 \times 10^{-25} \][/tex]
Next, we divide by the wavelength [tex]\( \lambda \)[/tex]:
[tex]\[ E = \frac{1.987821045 \times 10^{-25}}{3.45 \times 10^{-3}} \][/tex]
[tex]\[ E = 5.761798681 \times 10^{-23} \, \text{J} \][/tex]
So, the energy of the electrons with a wavelength of [tex]\( 3.45 \times 10^{-3} \)[/tex] meters is approximately:
[tex]\[ E \approx 5.76 \times 10^{-23} \, \text{J} \][/tex]
This is the kinetic energy per electron.
