Question

Question #438c2

Answer 1

`sf(1.3xx10^(-19)color(white)(x)J` or `sf(0.82color(white)(x)"eV")`

Explanation:

The incident photon strikes the electron with sufficient energy to ionise it. Any excess energy appears as the kinetic energy of the electron.

`sf(hf=KE+w)`

w is the work function of the metal and is the energy required to just remove the electron.

`:.` `sf(KE=hf-w)`

Converting eV to Joules:

`sf(w=2.3xx1.6xx10^(19)=3.68xx10^(-19)color(white)(x)J)`

Since we are given the wavelength `sf(lambda)` we can write:

`sf(KE=(hc)/(lambda)-w)`

`:.` `sf(KE=(6.63xx10^(-34)xx3.00xx10^(8))/(400xx10^(-9))-3.68xx10^(-19)color(white)(x)J)`

`sf(KE=(4.9925-3.68)xx10^(-19)color(white)(x)J)`

`sf(KE=1.312xx10^(-19)color(white)(x)J)`

If you want this in electron volts you divide by the electronic charge:

`sf(KE=(1.312xxcancel(10^(-19)))/(1.6xxcancel(10^(-19)))=0.82color(white)(x)"eV")`