Photoelectrons from a material with a binding energy of 271

Photoelectrons from a material with a binding energy of 2.71 eV are ejected by 425 nm photons. Once ejected, how long does it take these electrons to travel 2.30 cm to a detection device?

Solution

binding energy = 2.71 eV
= (2.71*1.6*10^-19) J
= 4.336*10^-19 J

Energy of 425 nm = h*c/lambda
= (6.626*10^-34 * 3*10^8) / (425*10^-9)
= 4.677*10^-19 J

Energy of released electron = 4.677*10^-19 J - 4.336*10^-19 J
= 3.41*10^-20 J

KE = 0.5*m*v^2
3.41*10^-20 = 0.5*9.1*10^-31 * v^2
v = 2.74*10^5 m/s

d = 2.3 cm = 2.3*10^-2 m

t = d/v
= (2.3*10^-2) / (2.74*10^5)
= 8.4*10^-8 s
Answer: 8.4*10^-8 s