Question

A monochromatic light beam is incident on a barium target that has a work function of 2.50 eV. If a potential difference of 1.00 V is required to turn back all the ejected electrons, what is the wavelength of the light beam?a) 355 nmb) 497 nmc) 744 nmd) 1.42 pme) none of those answers

Answer 1

Answer:

b) 497 nm

Explanation:

Given:

Work function, ϕ = 2.50 eV

Stopping Potential, V₀ = 1.00 eV

Charge of electron, e = 1.6 x 10⁻¹⁹ C

Speed of light, c = 3 x 10⁸ m/s

Planck's constant, h = 6.63 × 10⁻³⁴ Js = 4.14 × 10⁻¹⁵ eVs

Einstein photoelectric equation  is given by:

                                    $K.E_{max}$ = E - ϕ                 ----- (1)                  

$K.E_{max}$ is the maximum kinetic energy

E is the energy of the absorbed photons:  E = hf

ϕ is the work function of the surface:  ϕ = hf₀

The potential difference required to back all ejected electrons is called the Stopping Potential (V₀)

                       Stopping Potential (V₀) $= \frac{K.E. _{max}}{e}$

                                     $K.E. _{max} = eV_{o}$           -----(2)

Substituting (2) into (1)

                                        eV₀ = E - ϕ

                                      1.6 x 10⁻¹⁹ x 1 = E - 2.50

                                      E = 1.6 x 10⁻¹⁹ + 2.50

                                      E = 2.50 eV

But E = hf = hc/λ

                                     λ $= \frac{hc}{E}$

                                     λ = (4.14 × 10⁻¹⁵ × 3 x 10⁸) / 2.50

                                     λ = 1240 × 10⁻⁹ / 2.50

                                     λ = 496.8 × 10⁻⁹ m

                                     λ = 497 × 10⁻⁹ m  (approximately)

                                     λ = 497 nm