The work function for potassium is 2.24eV.

If potassium metal is illuminated with light of wavelength 310 nm,
find the maximum kinetic energy of the photoelectrons.

I have tried everything and still keep getting the wrong answer.

The speed of light is 3 × 108 and Planck’s constant is 6.62607 × 10−34
and the answer suppose to be in units of eV.

i used the equation MaxKE = h*c/ (wavelength x10^-9)

and then subtract (2.24 * 1.602x10^-19)

its not correct apparently, please help

i figured it out after all - thanks

use:

kmax = hf - e
kmax = h ( f becomes : c over lambda)- e

e = ( Evev * 1.6 *10exp -19)

To find the maximum kinetic energy of the photoelectrons, you need to use the equation:

MaxKE = energy of incident light - work function

Using the given information, we can calculate the energy of the incident light:

energy of incident light = (Planck's constant * speed of light) / wavelength

Let's plug in the values and solve the equation step by step:

1. Convert the wavelength from nm to meters:
wavelength = 310 nm * (1 m / 10^9 nm) = 310 × 10^-9 m

2. Calculate the energy of the incident light:
energy of incident light = (6.62607 × 10^-34 J s * 3 × 10^8 m/s) / (310 × 10^-9 m)
= (6.62607 × 3) / (310 × 10^-9) J
= 19.87821 × 10^-9 J
= 1.987821 × 10^-8 J

3. Convert the energy from joules to electron volts (eV):
1 eV = 1.602 × 10^-19 J
energy of incident light = (1.987821 × 10^-8 J) / (1.602 × 10^-19 J/eV)
≈ 124 eV

Now we can calculate the maximum kinetic energy of the photoelectrons:

MaxKE = 124 eV - 2.24 eV
= 121.76 eV

Therefore, the maximum kinetic energy of the photoelectrons is approximately 121.76 eV.

Please include your units for physical quantities. Numbers are not enough.