(a) [tex]2.5\cdot 10^{-6}eV[/tex]
The energy of a photon is given by:
[tex]E=\frac{hc}{\lambda}[/tex]
where
[tex]h=6.63\cdot 10^{-34}Js[/tex] is the Planck constant
[tex]c=3\cdot 10^8 m/s[/tex] is the speed of light
[tex]\lambda[/tex] is the wavelength
For the microwave photon,
[tex]\lambda=50.00 cm = 0.50 m[/tex]
So the energy is
[tex]E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.50 m}=4.0\cdot 10^{-25} J[/tex]
And converting into electronvolts,
[tex]E=\frac{4.0\cdot 10^{-25}J}{1.6\cdot 10^{-19} J/eV}=2.5\cdot 10^{-6}eV[/tex]
(b) [tex]2.5 eV[/tex]
For the energy of the photon, we can use the same formula:
[tex]E=\frac{hc}{\lambda}[/tex]
For the visible light photon,
[tex]\lambda=500 nm = 5 \cdot 10^{-7}m[/tex]
So the energy is
[tex]E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-7} m}=4.0\cdot 10^{-19} J[/tex]
And converting into electronvolts,
[tex]E=\frac{4.0\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.5 eV[/tex]
(c) [tex]2500 eV[/tex]
For the energy of the photon, we can use the same formula:
[tex]E=\frac{hc}{\lambda}[/tex]
For the x-ray photon,
[tex]\lambda=0.5 nm = 5 \cdot 10^{-10}m[/tex]
So the energy is
[tex]E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-10} m}=4.0\cdot 10^{-16} J[/tex]
And converting into electronvolts,
[tex]E=\frac{4.0\cdot 10^{-16}J}{1.6\cdot 10^{-19} J/eV}=2500 eV[/tex]
