Pieces of potassium (Φ = 3.68 × × 10–19 J) and sodium (Φ = 4.41 × × 10–19 J) metal are exposed to radiation of wavelength 260.0 nm.

Question

1. Which metal emits electrons with the greater velocity?
2. What is the velocity of the electrons from potassium?
3. What is the velocity of the electrons from sodium?

Answer 1

Φ=hv₀ KE_electron=hv-Φ v=c/λ

where Φ is work function of the metal, h is planks constant, v is frequncy, c is the speed of light, and λ is the wavelenght

for potassium:

v=(2.998E8)/260E-9 v=1.15E15 Hz

KE= (6.626E-34)(1.15E15)-3.68E-19

KE= 3.96E-19

KE=1/2mv² mass must be in kg v here is velocity not frequncy

3.96E-19=(1/2)(0.039098)v²

v=4.5E-9 m/s

for sodium:

KE= (6.626E-34)(1.15E-15)-4.41E-19

KE=3.21E-19

KE=1/2 mv²

3.21E-19=(1/2)(0.02299)v²

v=5.28E-9