Question

A certain metal with work function of Φ = 1.7 eV is illuminated in vacuum by 1.4 x 10-6 W of light with a wavelength of λ = 600 nm. 1)How many photons per second, N, are incident on the metal? N = photons per second Submit 2)What is KEmax, the maximum kinetic energy of the electron that is emitted from the metal? KEmax = eV

Answer 1

Answer:

1) n = 4.47*10^12 photons

2) K = 0.25 eV

Explanation:

This is a problem about the photoelectric effect.

1) In order to calculate the number of photons that impact the metal, you take into account the power of the light, which is given by:

$P=\frac{E}{t}=1.4*10^{-6}\frac{J}{s}$     (1)

Furthermore you calculate the energy of a photon with a wavelength of 600nm, by using the following formula:

$E_p=h\frac{c}{\lambda}$         (2)

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

h: Planck's constant = 6.626*10^-34 Js

λ: wavelength = 600*10^-9 m

You replace the values of the parameters in the equation (2):

$E_p=(6.626*10^{-34}Js)\frac{3*10^8m/s}{600*10^{-9}m}=3.131*10^{-19}J$

Next, you calculate the quotient between the power of the light (equation (1)) and the energy of the photon:

$n=\frac{P}{E_p}=\frac{1.4*10^{-6}J/s}{3.131*10^{-19}J}=4.47*10^{12}photons$

The number of photons is 4.47*10^12

2) The kinetic energy of the electrons emitted by the metal is given by the following formula:

$K=E_p-\Phi$     (3)

Ep: energy of the photons

Φ: work function of the metal = 1.7 eV

You first convert the energy of the photons to eV:

$E_p=3.131*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.954eV$

You replace in the equation (3):

$K=1.95eV-1.7eV=0.25eV$

The kinetic energy of the electrons emitted by the metal is 0.25 eV