(a) An electron with initial kinetic energy

Question

32

eV encounters a square barrier with height

41

eV and width

0.25

nm. What is the probability that the electron will tunnel through the barrier?

(b) A proton with the same kinetic energy encounters the same barrier. What is the probability that the proton will tunnel through the barrier?

Accepted Answer

Hey everyone. So this problem is dealing with quantum mechanics and tunneling probabilities. Let's see what it's asking us. We have a potential barrier that has a square geometry with a length of 0.32 nanometers in an energy of 36 electron volts. There are two parts to this problem. In the first part, an electron with an initial kinetic energy of 28 electron volts is shot towards the barrier and were asked to determine the tunneling probability of the electron. In part two, we're asked to determine the tunneling probability of a neutron that is shot towards the same barrier with the same initial kinetic energy uh as that of the electron. So our multiple choice answers are a 0.632 and 9.38 B 0.00026 and zero C. Both probabilities are zero or D both are 0.00026. So the key to solving this problem is we're calling the equation for our tunneling probability T which is equal to 16, multiplied by uppercase E or energy divided by U not multiplied by the quantity one minus E divided by you not multiplied by lowercase E or natural exponent raised to the power of negative two, multiplied by K or wave number multiplied by L our length. And so we have almost everything we need that was given to us from the problem as far as um our unknowns are variables here. So uppercase E our energy of the particle. So for part one, the electron is 28 ev our potential barrier, initial energy 36 ev and then K is our wave number which will solve four. And L is, was given as 0.32 nanometers which I will rewrite as 0.32 times 10 to the negative 9 m. So like I said, the only thing we need to solve for before we can plug into our main equation is this K or wave number. So we can recall that K is equal to the square root of two M divided by H bar squared multiplied by you not minus E. And so we do have everything we need to solve for K. So K is equal to two multiplied by the mass. And so I'll write this K sub E. So the mass of the electron we can recall is 9.11 times 10 to the negative 31 kg divided by H bar squared. So our reduced flanks constant or H bar we can recall is 1.055 times 10 to the negative 34 jewels multiplied by seconds. And that quantity is squared. And then again, on the numerator and the numerator of this term, we have, you know, minus E so 36 EV minus 28 EV. But we also need to recognize that we have electron volts in the numerator joules in the denominator. So we will have to use our conversion factor of 1.602 times 10 to the negative 19 joules her electron volt. And so from there, we can find that the wave number for our electron is 1.45 times 10 to the 10. And that's in units of one over meter per meter to the negative one. Recognizing that we're going to need to do the same thing for our neutron. We can solve for RK sub N for part two here. Now, how did we figure out how did we come to that conclusion? The only thing that is different between parts one and parts two. Part two of this problem is the particle. So part one is an electron, part two is a neutron, they have the same energy. So the only difference in them is their mass and the only place that the mass comes in into this equation is through this k term through this wave number. And so that is the only difference um between parts one and two. So while we're working with our wave numbers, let's solve four k of our neutron as well. So we can recall that the mass of our neutron is 1.67 times 10 to the negative 27 kg. And like I said, everything else remains the same. And so K sub NK for our neutron is 6.24 times 10 to the 11. And that's meters to the negative one. I might have misspoke earlier and said nucleus instead of neutron, but it is the particle is the neutron. OK. So now we have everything we need to plug into our tunneling probability equation. And so we'll do that first for part one for our electron. And so that is 16, multiplied by our energy 28 ev divided by you not 36 ev multiplied by the quantity one minus 28 EV divided by 36 ev multiplied by our natural exponent raised to negative two, multiplied by that K we solve for 1.45 times 10 to the 10 m of the negative one multiplied by L which the problem tells us is 0.32 times 10 to the negative 9 m. So in this part of the problem, I'm not changing electron volts to joules. I don't need to convert because they, the units cancel um in, in both of those terms. So that makes it one step easier plug that in and we get 2.6 times 10 to the negative four which we can rewrite as 0.00026. So that is the answer for part one. Similarly, for part two, the only thing that has changed is that K term because that's the only place that mass comes into play. So we can rewrite it almost exactly the same plugging in our wave number. For our neutron, we have 6.24 times 10 to the 11 m to the negative one. And that is equal to zero, depending on your calculator. You might get something, some value to the, you know, 10 to the negative 100 or plus. Um And it is essentially zero. It's such a small, small number. And so that is the answer to part two. And so when we go up to our multiple choice answers, we can see that answer choice and B is the correct answer for this problem. So that's all we have for this one. We'll see you in the next video.

Key Concepts

Quantum Tunneling

Barrier Penetration Probability

Mass and Energy Relationship

Watch next