Question

When ultraviolet light with a wavelength of 400 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is 1.10 eV.

What is the maximum kinetic energy K0 of the photoelectrons when light of wavelength 300 nm falls on the same surface?


Use h = 6.63 x 10−34J*s for Planck's constant and c = 3.00 x 108m/s for the speed of light and express your answer in electron volts.

Answer 1

Answer:

$K_{max}$ = 1.47 eV

Explanation:

This is an exercise that we can solve using the photoelectric effect, which stable that photons can be treated as particles and collide with the material, the process is described by the expression

      $K_{max}$ + Φ = h f

Where Kmanx is the maximum energy of the torn electrons, Φ is the work function of the material and hf is the energy of the photons

With the initial data we calculate the job function

We use the relationship of wave velocity with wavelength and frequency

    c = λ f

    f = c / λ

    f = 3 10⁸/400 10⁻⁹

    f = 7.5 10¹⁴ Hz

Let's reduce the magnitude to the SI system

    $K_{max}$ = 1.10 eV (1.6 10⁻¹⁹ J / 1 ev = 0.6875 10⁻¹⁹ J

    Φ = h f - $K_{max}$

    Φ = 6.63 10⁻³⁴ 7.5 10¹⁴ - 0.6875 10⁻¹⁹

    Φ = 4.9725 10⁻¹⁹ - 0.6875 10⁻¹⁹

    Φ = 4.275 10⁻¹⁹ J

Now let's calculate the frequency of the other wavelength

     f = c / λ₂

     f = 3 10⁸/300 10⁻⁹

     f = 1 10¹⁵ Hz

We calculate

    $K_{max}$ = hf - Φ

    $K_{max}$ = 6.63 10⁻³⁴ 1 10¹⁵ - 4.275 10⁻¹⁹

    $K_{max}$ = 6.63 10⁻¹⁹ - 4,275 10⁻¹⁹

    $K_{max}$ = 2,355 10⁻¹⁹ J

    $K_{max}$ = 2,355 10⁻¹⁹ j (1 eV / 1.6 10⁻¹⁹ j)

   $K_{max}$ = 1.47 eV