Final answer:
The binding energy in eV of electrons in aluminum, if the longest-wavelength photon that can eject them is 304 nm, is approximately 4.09 eV. Option a is the correct answer.
Step-by-step explanation:
To calculate the binding energy of electrons in aluminum, we need to use the equation:
E = hc/λ - Φ
where:
E is the binding energy
h is Planck's constant (6.63 x 10^-34 J·s)
c is the speed of light (3 x 10^8 m/s)
λ is the wavelength of the photon (304 nm converted to meters)
Φ is the work function (the minimum energy required to remove an electron from the metal)
First, we convert the wavelength to meters:
λ = 304 nm = 304 x 10^-9 m
Next, we calculate the binding energy:
E = (6.63 x 10^-34 J·s x 3 x 10^8 m/s) / (304 x 10^-9 m) - Φ
Since the longest-wavelength photon can eject the electrons, it means the binding energy is equal to the energy of the photon:
E = 6.63 x 10^-19 J - Φ
To convert the binding energy to eV, we divide by the elementary charge:
E (eV) = (6.63 x 10^-19 J - Φ) / (1.6 x 10^-19 C)
Given that Φ for aluminum is approximately 4.08 eV, we plug in the values:
E (eV) = (6.63 x 10^-19 J - 4.08 eV * 1.6 x 10^-19 C)
Simplifying this equation gives us:
E (eV) ≈ 4.09 eV
Therefore, the correct option is a) 4.09 eV.