If the minimum (threshold) energy for the emission of an electron from the surface of a particular metal is 3.90 X 10-19 J, what is the kinetic energy of an electron emitted from the surface of the metal when it is irradiated by electromagnetic radiation of wavelength 213 nanometers?

Planck's constant = 6.626 x 10-34 Js
Speed of electromagnetic radiation = 3.0 x 108 ms-1

Kinetic energy of the photoelectron =

I am not sure what steps to use to solve this question.

v is1/2 mv squared 2

To solve this question, you need to use the equation for the energy of a photon, which is given by:

E = hc/λ

Where:
E is the energy of the photon
h is Planck's constant (6.626 x 10^-34 Js)
c is the speed of electromagnetic radiation (3.0 x 10^8 m/s)
λ is the wavelength of the electromagnetic radiation in meters

In order to find the kinetic energy of the photoelectron, you need to determine the energy of the photon first. Then you can subtract the threshold energy from the photon energy to find the kinetic energy.

Here are the steps to solve the question:

1. Convert the given wavelength from nanometers to meters:
213 nanometers = 213 x 10^-9 meters

2. Use the equation E = hc/λ to find the energy of the photon:
E = (6.626 x 10^-34 Js * 3.0 x 10^8 m/s) / (213 x 10^-9 meters)

3. Calculate the energy of the photon.

4. Subtract the threshold energy (3.90 x 10^-19 J) from the energy of the photon calculated in step 3.

5. The result obtained in step 4 is the kinetic energy of the emitted photoelectron.

Now, you can plug in the given values and follow these steps to solve for the kinetic energy.