Answer:
a. 4.733 × 10⁻¹⁹ J = 2.954 eV b i. yes ii. 0.054 eV = 8.651 × 10⁻²¹ J
Explanation:
a. Find the energy of the incident photon.
The energy of the incident photon E = hc/λ where h = Planck's constant = 6.626 × 10⁻³⁴ Js, c = speed of light = 3 × 10⁸ m/s and λ = wavelength of light = 420 nm = 420 × 10⁻⁹ m
Substituting the values of the variables into the equation, we have
E = hc/λ
= 6.626 × 10⁻³⁴ Js × 3 × 10⁸ m/s ÷ 420 × 10⁻⁹ m
= 19.878 × 10⁻²⁶ Jm ÷ 420 × 10⁻⁹ m
= 0.04733 × 10⁻¹⁷ J
= 4.733 × 10⁻¹⁹ J
Since 1 eV = 1.602 × 10⁻¹⁹ J,
4.733 × 10⁻¹⁹ J = 4.733 × 10⁻¹⁹ J × 1 eV/1.602 × 10⁻¹⁹ J = 2.954 eV
b. i. Is this energy enough for an electron to leave the atom
Since E = 2.954 eV is greater than the work function Ф = 2.9 eV, an electron would leave the atom. So, the answer is yes.
ii. What is its maximum energy?
The maximum energy E' = E - Ф = 2.954 - 2.9
= 0.054 eV
= 0.054 × 1 eV
= 0.054 × 1.602 × 10⁻¹⁹ J
= 0.08651 × 10⁻¹⁹ J
= 8.651 × 10⁻²¹ J
