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.