To eject electrons from the surface of potassium metal, 222kJ/mol of electrons are needed.

a) What minimum frequency of light is needed to supply the required energy?
b) What is the wavelength of this light?
c) If potassium is irradiated with light having wavelength of 350 nm, what is the maximum E(sub K) of an emitted electron?
d) What maximum speed in m/2 will the electrons in part (c) have?

Divide 222 kJ/mol by 6.022E-23 which gives you per photon, then convert to J/photon.

a. E = h*frequency
b. c = frequency*wavelength
c. E = hc/wavelength
d. E = 1/2 mv^2

The number you have from 222,000/6.022E23 (that's not -23 and I've made it Joules by converting kJ to J) is called the work function. For part a,

E = h*frequency-work function. I forgot the work function part.

a) To find the minimum frequency of light needed to supply the required energy, we can use the equation:

E = h * f

Where E is the energy required to eject electrons from the surface of potassium metal (222 kJ/mol), h is Planck's constant (6.626 x 10^(-34) J·s), and f is the frequency of light we are looking for.

To convert the energy from kJ to J, we multiply by 1000:

E = 222 kJ/mol * 1000 J/kJ = 222,000 J/mol

Now, we need to convert J/mol to J to find the energy required per photon. This can be done by dividing by Avogadro's number, which is 6.022 x 10^23 mol^(-1):

Energy per photon = 222,000 J/mol / (6.022 x 10^23 mol^(-1))

Next, we can rearrange the equation E = h * f to solve for f:

f = E / h

f = (222,000 J/mol) / (6.022 x 10^23 mol^(-1) * 6.626 x 10^(-34) J·s)

By plugging in the values and calculating, we can find the minimum frequency of light needed.

b) To find the wavelength of this light, we can use the equation:

c = λ * f

Where c is the speed of light (3.0 x 10^8 m/s), λ is the wavelength of light, and f is the frequency we calculated in part (a).

By rearranging the equation, we can solve for λ:

λ = c / f

Now, we can plug in the values and calculate the wavelength of the light.

c) To find the maximum kinetic energy (E(sub K)) of an emitted electron when potassium is irradiated with light having a wavelength of 350 nm, we can use the equation:

E(sub K) = h * f - φ

Where E(sub K) is the maximum kinetic energy, h is Planck's constant (6.626 x 10^(-34) J·s), f is the frequency of light (which we can calculate from the given wavelength in part (b)), and φ is the work function, which represents the energy required to overcome the binding energy of an electron in an atom or material.

To find φ for potassium, we can use the given information that 222 kJ/mol of energy is needed to eject electrons from the surface of potassium metal. We can convert this to J/mol by multiplying by 1000:

φ = 222 kJ/mol * 1000 J/kJ = 222,000 J/mol

Now, we can calculate the frequency of light using the wavelength given (350 nm) and then plug in the values to find the maximum kinetic energy.

d) To find the maximum speed in m/s that the electrons from part (c) will have, we can use the kinetic energy equation:

E(sub K) = (1/2) * m * v^2

Where E(sub K) is the maximum kinetic energy, m is the mass of an electron (9.11 x 10^(-31) kg), and v is the maximum speed we are looking for.

We can rearrange the equation to solve for v:

v = sqrt((2 * E(sub K)) / m)

By plugging in the values and calculating, we can find the maximum speed of the electrons.