Pieces of potassium (Φ = 3.68 ×× 10–19 J) and sodium (Φ = 4.41 ×× 10–19 J) metal are exposed to radiation of wavelength 350.0 nm.

Question

Pieces of potassium (Φ = 3.68 ×× 10^–19 J) and sodium (Φ = 4.41 ×× 10^–19 J) metal are exposed to radiation of wavelength 350.0 nm.

What is the velocity of the electrons from potassium? What is the velocity of the electrons from potassium?

Answer 1

Pieces of potassium (Φ = 3.68 ×× 10^–19 J) and sodium (Φ = 4.41 ×× 10^–19 J) metal are exposed to radiation of wavelength 350.0 nm.

What is the velocity of the electrons from potassium? What is the velocity of the electrons from potassium?

Is not hf = φ_o + E_{K ... ?}

If so, then just replace the kinetic energy with 1/2mv² and solve for v to find the escape velocity of the electron.

h = plancks const = 6.626 x10^-34 J*s

f = frequency in hz which is found using c/λ = f

Electron mass m_e = 9.109 ×10⁻³¹ Kilograms

hf = φ_o + E_K

6.626E^-34J*s * (3E⁸m/s / 350E^-9m) = 3.68E^-19J + 1/2 * 9.1E^-31kg * v²

6.626E^-34J*s * 8.57E¹⁴ 1/s ⁼

5.68E^-19 J - 3.68E^-19J =

√[2(2E^-19) J / 9.1E^-31kg] = v

662,993 m/s = v

6.63x10⁵ = v

I could be totally wrong!