Question

When a light of wavelength 570nm falls on the surface of potassium metal, electrons are ejected with a velocity of 6.4x10^4 m/s. What is the minimum energy required to remove an electron from potassium metal?

Answer 1

Answer:

E = hc/lembda

E = 2×10^-25/570×10^-9

calculate this to get value of E