(a) Light of wavelength 300 nm is incident on a metal that...

(a) Light of wavelength 300 nm is incident on a metal that has a work function of 1.4 eV. What is the maximum speed of the emitted electrons? (b) Repeat part (a) for light of wavelength 800 nm incident on a metal that has a work function of $1.6 \mathrm{eV} .$ (c) How would your answers to parts (a) and (b) vary if the light intensity were doubled?

Key Concepts

Light Intensity Impact

Intensity in the photoelectric effect determines the number of photons hitting the metal surface per unit time, thereby influencing the number of electrons emitted. However, the maximum kinetic energy (and thus the speed) of the ejected electrons is solely dependent on the energy of single photons and is independent of the light intensity, assuming the frequency remains constant.

Kinetic Energy and Electron Speed

Once an electron is emitted, its maximum kinetic energy is given by the difference between the photon's energy and the metal's work function (E_kin = E_photon - work function). This kinetic energy relates to the speed of the electron via the equation 0.5 m v^2, where m is the electron mass, allowing computation of the electron's maximum velocity.

Work Function

The work function is the minimum energy required to remove an electron from the surface of a metal. In the context of the photoelectric effect, it sets the threshold energy that the incident photon must exceed for electron emission to occur.

Photon Energy

Photon energy is determined by the equation E = hc/?, where h is Planck's constant, c is the speed of light, and ? is the wavelength of the incident light. This relationship shows that shorter wavelengths carry higher energy, which is crucial for ejecting electrons in the photoelectric effect.

Photoelectric Effect

This phenomenon involves the emission of electrons from a metal surface when light of sufficient energy strikes it. The process depends on the light's frequency (or wavelength) and the material's properties, such as its work function, rather than on the light's intensity.

Transcript

00:01 For this exercise, we have a light of wavelength 300 nanometers, shining on a certain metal plate whose work function phi is 1 .4 electron volts.

00:16 And we have to calculate the maximum velocity of the ejected electrons.

00:23 So for that, we're going to calculate the maximum kinetic energy k.

00:27 And according to the photoelectric formula we have that k equals h f minus phi f can be written as c over lambda and hc is twelve hundred and forty lambda is three hundred nanometers and hc is in is twelve hundred electron volt nanometers minus the work function and this equals 2 .73 electron volts and converting remember that one electron volt equals 1 .6 times 10 to the 9 to the minus 19 joules so this here is the same as 4 .37 times 10 to the minus 19 joules okay and we know that the kinetic energy equals mass times velocity squared over 2.

01:39 So the velocity can be obtained from the kinetic energy as velocity equals 2k over m.

01:52 And we have to remember that the max of the electron equals 9 .1 times 10 to the minus 31 kilograms.

02:04 So we can simply substitute all we have here.

02:11 It's going to be 2 times 4 .37 times 10 to the minus 19th over the mass there, 9 .1 times 10 to the minus 31.

02:28 And this here equals 9 .8 times 10 to the 5 meters per second.

02:38 So this is the velocity.

02:42 This is the answer to question a.

02:44 This is the maximum speed of the electrons.

02:49 Okay.

02:51 So for question b, the exercise asks us to calculate the same thing, the maximum velocity, but for a wavelength of 800 nanometers and a work function of 1 .6 instead of 1 .4 electron volts.

03:14 So we got to do the same thing...

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