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A copper cathode in a vacuum chamber is illuminated with light from a laser, causing electrons to be emitted from the surface of the metal.
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The light has a frequency of 1.80 times 10 to the 15th hertz.
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The maximum kinetic energy of the ejected electrons is 2.80 electron volts.
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What is the work function of copper?
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Use a value of 4.14 times 10 to the negative 15th electron volt seconds for the value of the Planck constant.
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Give your answer in electron volts to three significant figures.
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Okay, so in this exercise, we have a copper cathode.
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And let’s say that this is that cathode.
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It’s in a vacuum chamber being illuminated, which causes some electrons to be ejected from the copper.
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We’re told that the frequency of the incoming radiation, we’ll call it 𝐹, is 1.80 times 10 to the 15th hertz and that the maximum kinetic energy of the electrons ejected from the copper, we’ll call it 𝐾𝐸 sub 𝑚, is 2.80 electron volts.
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Recall that one electron volt is equal to the amount of energy a charge of one electron gets when it’s moved across the potential difference of one volt.
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Knowing all this, we want to solve for the work function of copper.
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To figure this out, we can recall a relationship between frequency, 𝐹, maximum kinetic energy, 𝐾𝐸 sub 𝑚, and the work function of a material, called 𝑊.
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That relationship states that Planck’s constant ℎ multiplied by the frequency of incoming radiation minus the work function of a material is equal to the maximum kinetic energy of ejected electrons.
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Since we want to solve for the work function 𝑊, we’ll rearrange this equation by subtracting 𝐾𝐸 sub 𝑚 from both sides and adding the work function 𝑊 to both sides.
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When we do that, we find that 𝑊 is equal to ℎ times 𝑓 minus 𝐾𝐸 sub 𝑚.
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Note that a work function 𝑊 is specific to a particular material.
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In our case, we’re solving for the work function of our copper cathode.
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To help us solve for it, we’re given the frequency 𝑓 of incoming radiation as well as the maximum kinetic energy of ejected electrons.
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We’re also told to treat Planck’s constant ℎ as 4.14 times 10 to the negative 15th electron volt seconds.
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So let’s substitute those three values into this equation.
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With those values plugged in, let’s take a look for a second at the units involved.
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In our first term, we have electron volt seconds multiplied by hertz.
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But hertz, the number of cycles performed per second, can be replaced with one over seconds.
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And when we do this, we see that in multiplying these two values together, that factor of seconds will cancel out.
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So now, we have one number in electron volts being subtracted from another number in electron volts.
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Therefore, our final answer will be in these desired units.
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And when we calculate the value of this expression to three significant figures, we find a result of 4.65 electron volts.
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This is the work function of copper.