Alight source of wavelength λ illuminates a metal with a work function (a. k.a., binding energy) of be=2.00 ev and ejects electrons with a maximum ke=4.00 ev. a second light source with double the wavelength of the first ejects photoelectrons with what maximum kinetic energy?

Explanation:

The described situation is the photoelectric effect, which consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.  

If we consider the light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.

This is what Einstein proposed:  

Light behaves like a stream of particles called photons with an energy  :

(1)  

So, the energy of the incident photon must be equal to the sum of the Work function of the metal and the kinetic energy of the photoelectron:  

(2)  

Where is the minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and its value depends on the metal.  

In this case  and

So, for the first light source of wavelength , and  applying equation (2) we have:

  (3)  

  (4)  

Now, substituting (1) in (4):  

(5)  

Where:  

is the Planck constant

is the frequency  

Now, the frequency has an inverse relation with the wavelength

:  

(6)  

Where is the speed of light in vacuum  

Substituting (6) in (5):  

(7)  

Then finding :  

(8)  

We obtain the wavelength of the first light suorce :  

  (9)

Now, we are told the second light source  has the double the wavelength of the first:

  (10)

Then:   (11)

Knowing this value we can find :

  (12)

  (12)

  (13)

Knowing the value of and , and knowing we are working with the same work function, we can finally find the maximum kinetic energy for this wavelength:

(14)  

(15)  

 This is the maximum kinetic energy for the second light source

1/0.14nm = 10^10/1.4 a lot

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.