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Take gravitational constant g = 10 m/s2.
1) The work function for calcium is 2.9 ev. If we shine on the surface made of calcium with a
light of 420 nm wavelength, answer the following questions:
a) Find the energy of the incident photon.
b) Is this energy enough for an electron to leave the atom and if so what is its
maximum energy?

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

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