Question Video: Using Electron Wavelength and Kinetic Energy to Solve for an Unknown Value Physics

The wavelength πœ† of an electron that has a kinetic energy of 𝐸 can be expressed as πœ† = π‘˜(β„Ž/𝐸), where β„Ž is the Planck constant and π‘˜ is a variable. Which of the following is equal to π‘˜? [A] Half the velocity of the electron. [B] Half the momentum of the electron. [C] Half the mass of the electron. [D] None of the answers are correct.

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Video Transcript

The wavelength πœ† of an electron that has a kinetic energy of 𝐸 can be expressed as πœ† equals π‘˜ times β„Ž divided by 𝐸, where β„Ž is the Planck constant and π‘˜ is a variable. Which of the following is equal to π‘˜? (A) Half the velocity of the electron. (B) Half the momentum of the electron. (C) Half the mass of the electron. (D) None of the answers are correct.

We’re considering here this equation, where πœ† is the wavelength of an electron with kinetic energy 𝐸 and β„Ž is the Planck constant. We want to figure out which one of our answer options describes this variable π‘˜. One way to eliminate some of these options is to think about the units on either side of our equation. Because this is an equation, the units on the left must match those on the right.

If we work in SI base units where possible, then the units of wavelength will be meters, the units of the Planck constant β„Ž will be joules times seconds, and the units of the kinetic energy of our electron will be joules. We can therefore write this equation, which is just an equation of units. On the left, we have meters, the units of wavelength, and then we have the units of π‘˜, whatever those are, multiplied by joules times seconds divided by joules.

We can see that the units of joules in numerator and denominator will cancel out. And this means that the units of this variable π‘˜ multiplied by seconds must be equal to meters. This equation then requires that the units of π‘˜ are meters per second. That way this factor of seconds in the numerator will cancel with this factor in the denominator. And we’ll arrive at a simplified equation, which says that meters is equal to meters, which is true. Since the units of π‘˜ are some distance, in this case in units of meters, divided by some time, in this case in units of seconds, we know that π‘˜ must represent a speed or a velocity.

Looking at our answer options, we know that option (B) is not correct. π‘˜ can’t be a momentum because the units of momentum are not the same as those of velocity. For the same reason, option (C) can’t be correct either. The units of mass are not the same as those of velocity. At this point, we’re inclined towards answer option (A). Notice though that this option has this somewhat strange phrasing β€œHalf the velocity of the electron.” So far, we’ve identified that π‘˜ is a velocity, but we haven’t yet found that it’s half of anything. To see whether that’s the case or not, we’ll need to approach this question from a different perspective.

Recall that we’re considering an electron that has a wavelength πœ†. That wavelength is described by the de Broglie equation. And this equation says that the wavelength of some particle is equal to Planck’s constant β„Ž divided by the particle’s momentum 𝑝. The momentum 𝑝 of an object is equal to its mass times its velocity. We have then that the de Broglie wavelength of some object equals Planck’s constant divided by the object’s mass times its velocity.

Clearing some space to work, we can apply the de Broglie wavelength of the electron to this given equation. That is, in addition to saying that πœ† is equal to π‘˜ times β„Ž divided by 𝐸, we can say that’s equal to β„Ž divided by π‘š times 𝑣. Focusing on just this equation, we can see that the Planck constant β„Ž is common to both sides, and therefore we can cancel it out.

We have then that our variable π‘˜ divided by the kinetic energy of the electron equals one divided by the electron’s mass times its velocity. Let’s now recall that an object’s kinetic energy β€” we’ll refer to it as 𝐸 β€” is equal to one-half that object’s mass times its velocity squared. Making this substitution, we find that π‘˜ divided by one-half π‘šπ‘£ squared equals one over π‘š times 𝑣.

If we multiply both sides of this equation by π‘š times 𝑣, then on the left-hand side, one factor of π‘š and one factor of 𝑣 cancel from numerator and denominator. And on the right-hand side, both of these factors cancel out completely. Our equation simplifies to π‘˜ divided by one-half times 𝑣 is equal to one. Multiplying both sides by one-half times the velocity 𝑣, that factor cancels on the left, and we find the final simplified result that π‘˜ equals 𝑣 divided by two.

This then is what answer option (A) is speaking to when it describes π‘˜ as half the velocity of the electron. π‘˜ is literally equal to 𝑣 divided by two. We choose then answer option (A). The variable π‘˜ is equal to half the velocity of the electron.

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