Answer:
(a) To calculate the energy in electron volts of an electron with a de Broglie wavelength of 400 nm, we can use the de Broglie wavelength equation:
λ = h / p
where λ is the de Broglie wavelength, h is Planck's constant (6.626 x 10^-34 J s), and p is the momentum of the particle. We can rearrange this equation to solve for the momentum:
p = h / λ
Plugging in the given de Broglie wavelength, we get:
p = (6.626 x 10^-34 J s) / (400 x 10^-9 m)
= 1.6565 x 10^-24 kg m/s
To calculate the kinetic energy of the electron, we can use the formula:
KE = p^2 / (2m)
where m is the mass of the electron (9.109 x 10^-31 kg). Plugging in the momentum we just calculated, we get:
KE = (1.6565 x 10^-24 kg m/s)^2 / (2 x 9.109 x 10^-31 kg)
= 1.423 x 10^-17 J
Finally, we can convert this energy from joules to electron volts (eV) using the conversion factor 1 eV = 1.602 x 10^-19 J:
KE = 1.423 x 10^-17 J / (1.602 x 10^-19 J/eV)
= 88.8 eV
Therefore, the energy in electron volts of an electron with a de Broglie wavelength of 400 nm is 88.8 eV.
(b) To calculate the energy in electron volts of a photon with a wavelength of 400 nm, we can use the formula:
E = hc / λ
where E is the energy of the photon, h is Planck's constant, c is the speed of light (299,792,458 m/s), and λ is the wavelength of the photon. Plugging in the given wavelength, we get:
E = (6.626 x 10^-34 J s) x (299,792,458 m/s) / (400 x 10^-9 m)
= 4.965 x 10^-19 J
Finally, we can convert this energy from joules to electron volts using the conversion factor 1 eV = 1.602 x 10^-19 J:
E = 4.965 x 10^-19 J / (1.602 x 10^-19 J/eV)
= 3.10 eV
Therefore, the energy in electron volts of a photon with a wavelength of 400 nm is 3.10 eV.
Step-by-step explanation: