Question

Find the least frequency of incident light that will knock electrons out of the surface of a metal with a work function of 3eV.​

Answer 1

Answer:

The frequency must be: $725\,\,10^{12} \,Hz$

Explanation:

If the work function of the metal ($\phi$) is 3 eV, then we can use the formula for the kinetic energy of an ejected electron:

$KE= h*f-\phi$

considering for the minimum KE = 0, and using the Plank constant h in eV s as: 4.14 * 10 ^(-15) eV s, to solve for the frequency:

$h*f=\phi\\4.14*10^{-15} * f = 3\\f=3*10^{15} /4.14\,\,\frac{1}{s} \\f=725\,10^{12} \,Hz$