Question

In order to comply with the requirement that energy be conserved, Einstein showed in the photoelectric effect that the energy of a photon (h) absorbed by a metal is the sum of the work function (), the minimum energy needed to dislodge an electron from the metal's surface, and the kinetic energy (Ek) of the electron: h = + Ek. When light of wavelength 357.4 nm falls on the surface of potassium metal, the speed (u) of the dislodged electron is 5.8×105 m/s. What is (in kJ/mol) of potassium?

Answer 1

Complete Question

In order to comply with the requirement that energy be conserved, Einstein showed in the photoelectric effect that the energy of a photon (h) absorbed by a metal is the sum of the work function (
`\phi`), the minimum energy needed to dislodge an electron from the metal's surface, and the kinetic energy (Ek) of the electron:
`hf = \phi +Ek`. When light of wavelength 357.4 nm falls on the surface of potassium metal, the speed (u) of the dislodged electron is 5.8×105 m/s. What is work function (in kJ/mol) of potassium?

Answer:

The work function is
`\phi_m = 242.195 \ KJ /mol`

Step-by-step explanation:

From the question we are told that

`hf = \phi +Ek`

Where h is the Planck's constant which has a constant value of
`h = 6.626 *10^(-34) J \cdot s`

Ek is the kinetic energy which is mathematically represented as

`Ek = (1)/(2) m v^2`

Where m is the mass of electron with a constant value

`m = 9.11 *10^(-31) \ kg`

The wavelength is
`\lambda = 357.4 nm = 357.4 *10^(-9) \ m`

The speed of the electron is
`v = 5.8*10^(5) m/s`

This photoelectric effect can be mathematically defined as

f is the frequency of the incident light which is mathematically represented as
`f = (c)/(\lambda )`

`\phi` is the work function

So we have

`(hc)/(\lambda ) = \phi + (1)/(2) mv^2`

substituting value

`(6.626 *10^(-34)* 3*10^8)/(357.4 *10^(-9)) = \phi + (1)/(2) * (9.11*10^(-31)) * (5.8*10^(8)) ^2`

`5.5618*10^(-19) = \phi + 1.54*10^(-14)`

=>
`\phi = 4.021*10^(-19) J`

For one mole of one mole the work function is

`\phi _m = \phi * N`

Where N is the avogadro's constant with a value

`N = 6.022*10^(23)`

So

`\phi_m = 4.021*10^(-19) * 6.022 *10^(23)`

`\phi_m = 242194.90 J /mol`

now in KJ

`\phi_m = 242.195 \ KJ /mol`