Question

EM radiation of an unknown wavelength falls at a normal incidence upon a piece of photoemissive material with a work function of 2.80eV. The max kinetic energy of ejected electrons is 1.79eV. What was the wavelength of the incident radiation? What portion of the EM spectrum does this fall into?

Answer 1

Answer:

  λ = 2.7 10⁻⁷ m   It is in the ultraviolet region

Explanation:

This is a photoelectric effect type experiment, where the photons behave like particles, in this case using energy conservation. The energy of the incident photon is equal to the energy to draw an electron (work function) plus the kinetic energy of the emitted particles

      E = Φ + K

 

Where E is the photon energy, Φ the work function and K the kinetic energy

Let's calculate E

     E = 280 + 1.79

     E = 4.59 eV

     E = 4.59 eV (1.6 10⁻¹⁹ J / 1 eV)

     E = 7.344 10⁻¹⁹ J

We use Planck's relationship

     E = h f

     f = E / h

     f = 7.344 10⁻¹⁹ / 6.63 10⁻³⁴

     f = 1,108 10¹⁵ Hz

  Now we use the equation of the speed of light

     c = λ f

     λ = c / f

     λ = 3 10⁸ / 1,108 10¹⁵

     λ = 2.7 10⁻⁷ m

Let's take that wavelength to nm

      λ = 2.7 10-7 (10⁹ nm / 1 m) = 2.7 10²

      λ= 2700 nm

It is in the ultraviolet region