Answer:
The frequency must be: [tex]725\,\,10^{12} \,Hz[/tex]
Explanation:
If the work function of the metal ([tex]\phi[/tex]) is 3 eV, then we can use the formula for the kinetic energy of an ejected electron:
[tex]KE= h*f-\phi[/tex]
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:
[tex]h*f=\phi\\4.14*10^{-15} * f = 3\\f=3*10^{15} /4.14\,\,\frac{1}{s} \\f=725\,10^{12} \,Hz[/tex]
