Question #438c2

1 Answer
Mar 12, 2017

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

Explanation:

theoraclemachine.net

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")#