The work function of a photoelectrode is 4.73 electron volts. Find the maximum velocity of photoelectrons ejected from the photoelectrode by 80.0-nanometre radiation.
We’re told in this statement that the work function of the photoelectrode is 4.73 electron volts. We’ll call that value 𝜙. We’re also told that radiation of 80.0-nanometre wavelength is used to eject photoelectrons from the photoelectrode; we’ll call that value 𝜆. We want to solve for the maximum velocity of ejected photoelectrons, what we’ll call 𝑉 sub max.
Let’s begin by drawing a picture of our situation. In this scenario, we have a surface, which we shine light of wavelength 𝜆 onto. The light is sufficiently energetic to dislodge electrons from the surface, given the work function of our particular surface which expresses how much energy needs to be overcome before it will release an electron and wavelength of the radiation we’re using.
We want to figure out how fast the dislodged electrons would move. There is a mathematical relationship that describes the energy of an ejected photoelectron. That relationship says that the photoelectron energy ℎ times 𝑓 equals the electron’s kinetic energy plus the work function 𝜙 that must be overcome for the electron to be ejected. We recall that kinetic energy is equal to one-half an object’s mass times the square of its speed.
And in this problem, we’ll assume that Planck’s constant ℎ is exactly 6.626 times 10 to the negative 34th joule-seconds. When we apply this relationship for electron energy to our situation, we see that ℎ times 𝑓 equals one-half the mass of the electron 𝑚 sub 𝑒 times the electron’s maximum speed 𝑉 sub max squared plus 𝜙. We’ll assume the electron’s mass is exactly 9.1 times 10 to the negative 31st kilograms. And since our incident radiation is a wave moving at the speed of light 𝑐, recalling that wave speed equals frequency times wavelength, which means that frequency equals speed over wavelength, we can substitute that expression in for 𝑓.
Now, we have an equation all in terms of known quantities or constant values. Now, we want to rearrange it to solve for 𝑉 sub max. When we subtract five from both sides of the equation, multiply both sides of the equation by two, divide by 𝑚 sub 𝑒, and then take square roots of both sides of the equation, we find that 𝑉 sub max equals the square root of two over 𝑚 sub 𝑒 times the quantity ℎ𝑐 over 𝜆 minus 𝜙.
We can now plug in for the variables in this equation. With all these values plugged in, there are two things to note about this expression. First, we’ve converted our wavelength originally given in nanometres into units of metres to make it consistent with the units of the rest of our values. And speaking of unit consistency, we’ll want to convert the energy of the work function given in electron volts to a value in units of joules.
To do that, we’ll multiply this value by the conversion factor from joules to electron volts. With that done, we’re now ready to enter these values on our calculator and compute 𝑉 sub max. When we do, we find that to three significant figures 𝑉 sub max is 1.95 times 10 to the sixth metres per second. That’s the maximum velocity at which an electron would be ejected from this surface.