Final answer:
The wavelength of the photons that eject 0.100-eV electrons from potassium is 1240 nm, which is in the infrared range and not visible to the human eye.
Step-by-step explanation:
To find the wavelength of photons that eject 0.100-eV electrons from potassium, we can use the equation:
λ = hc/E
Where λ is the wavelength of the photons, h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (3.00 x 10^8 m/s), and E is the energy required to eject the electron.
First, we need to convert the energy from eV to joules. 1 eV is equal to 1.6 x 10^-19 J.
So, the energy required to eject the electron is 0.100 eV x 1.6 x 10^-19 J/eV = 1.6 x 10^-20 J.
Now we can calculate the wavelength:
λ = (6.626 x 10^-34 J s x 3.00 x 10^8 m/s) / (1.6 x 10^-20 J) = 1.24 x 10^-6 m = 1240 nm
The wavelength of the photons that eject 0.100-eV electrons from potassium is 1240 nm, which is in the infrared range and not visible to the human eye.