Question

When light of a frequency (1.1x10¹⁵) s^-1 shines on the surface of aluminum metal, electrons are ejected with a maximum kinetic energy of (4.900x10^-19) J. Calculate the binding energy of electrons to aluminum metal.

Answer 1

Solution:

According to the theory of the photoelectric effect, the photon energy is partly used to break the electron out of the material and the remainder goes to the ejected electron kinetic energy. This, the binding energy $BE$ of electrons of aluminium metal will be the energy of the photon $h\nu$ (where $h$ is the Planck's constant 6.626·10$^{-34}$ (J·s) and $\nu$ is the frequency of the light) minus the maximum kinetic energy $KE$ :

$BE = h\nu - KE$

$BE = 6.626·10^{-34}·1.1·10^{15} (J)-4.9·10^{-19} (J)$

$BE = 2.393·10^{-19} (J)$

Answer: the binding energy of electrons to aluminum metal is 2.393·10^-19 J.