Question

If photons with a wavelength of 23.7 nm hits a metal surface with a binding energy of 496 kJ/mol, what would be the MOST LIKELY speed of the ejected electrons?

Does anyone know how to go about solving this? I'm kind of not sure how to even start.

Answer 1

Binding energy = 496 KJ/mol

binding energy per atom = 496 KJ / (6.022*10^23)

= 8.236*10^-22 KJ

= 8.236*10^-19 J

now find the energy of photon:

lambda = 23.7 nm = 2.37*10^-8 m

use:

E = h*c/lambda

=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(2.37*10^-8 m)

= 8.387*10^-18 J

energy of ejected electrons = E of photon - Eo

= 8.387*10^-18 J - 8.236*10^-19 J

= 7.563*10^-18 J

mass of electron = 9.11*10^-31 Kg

use:

KE = 0.5*m*v^2

7.563*10^-18 = 0.5*9.11*10^-31*v^2

v = 4.07*10^6 m/s

Answer: 4.07*10^6 m/s