Answer:
The binding energy of electrons in aluminum, if the longest-wavelength photon that can eject them is 304 nm, is 4.086 x 10^11 eV.
Step-by-step explanation:
To calculate the binding energy of electrons in aluminum, we can use the equation for the energy of a photon with a given wavelength.
The energy (E) of a photon is given by:
E = hc / λ
Where:
E = energy of the photon (in joules)
h = Planck's constant (approximately 6.626 x 10^-34 J·s)
c = speed of light (approximately 3.00 x 10^8 m/s)
λ = wavelength of the photon (in meters)
First, we need to convert the given wavelength from nanometers to meters:
λ = 304 nm = 304 x 10^-9 m
Now, we can calculate the energy of the photon:
E = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (304 x 10^-9 m)
E ≈ (1.9878 x 10^-25 J) / (304 x 10^-9 m)
Now, let's convert the energy from joules to electronvolts (eV). 1 eV is equal to 1.602 x 10^-19 joules.
E (in eV) ≈ (1.9878 x 10^-25 J) / (304 x 10^-9 m) * (1 eV / 1.602 x 10^-19 J)
E (in eV) ≈ (1.2422 x 10^14 eV) / (304 x 10^-9 m)
E (in eV) ≈ 4.086 x 10^11 eV
So, the binding energy of electrons in aluminum, if the longest-wavelength photon that can eject them is 304 nm, is 4.086 x 10^11 eV.