Question

Electrons are emitted form a certain metal with a maximum kinetic energy of 2 eV when 6-eV photons are inicident on its surface. What is the maximum kinetic energy of electrons emitted if photons of twice the wavelength are inicdent on this metal? Explain your answer

Answer 1

The energy of the light is not higher than the work function. Then, the electrons are not emitted.

Step-by-step explanation:

In order to calculate the maximum kinetic energy of the electrons for photons of twice the wavelength of the light, you first calculate the wavelength of photons with energy of 6eV. You use the following formula:

`E_p=h(c)/(\lambda)` (1)

c: speed of light = 3*10^8 m/s

λ: wavelength of the light

h: Planck's constant in eV.s = 4.135*10^-15 eV.s

E: energy of the photons = 6eV

You solve the equation (1) for λ:

`\lambda=(hc)/(E)=((4.135*10^(-15)eV)(3*10^8m/s))/(6eV)\\\\\lambda=2.06*10^(-7)m`

Next, you calculate the energy of photons with twice the wavelength:

`E_p'=h(c)/(2\lambda)=(4.135*10^(-15)eV)(3*10^8m/s)/(2(2.06*10^(-7)m))\\\\E_p'=3.0eV`

Next, you calculate the work function of the metal by using the equation for the photo electric effect:

`K=E_p-\Phi` (2)

Ф: work function

Ep: energy of the photons = 6eV

K: kinetic energy of emitted electrons = 2eV

You solve for Ф:

`\Phi=E_p-K=6eV-2eV=4eV` (3)

Finally, you calculate the kinetic energy of the emitted electron by the metal when the light with energy Ep' is used:

`K=E_P'-\Phi=3.0eV-4eV`

It is clear that for a light with energy 3.0eV has an energy lower than the work function of the metal, then, the electrons are not emmited by the metal