# Confused about quantum tunneling through 200V not eV

In summary: Energy and voltage (potential difference, p.d.) are different types of quantity. But they are related (by E = qV).If an electron passes though a p.d. of 1volt, it gains (or loses) an amount of potential energy; this amount of energy is called an electron-volt (eV). Converting between ##\Delta U## and ##\Delta V##, it's often just adding or removing an "e" from the unit. So that's why it may seem like volts and electron-volts are interchanged willy-nilly, but that's not really happening.An electron-volt is (a very small) unit of energy and
Homework Statement
How big of a fraction of a 50 eV beam is transmitted through a 1nm wide, 200V barrier?
Relevant Equations
Shrödinger equations
I have the equations to calculate transmission probability, my problem is that the barrier is given in Volts not electron volts.

$$200V = e \cdot 200 eV = 3.2 \cdot 10^{-17} eV$$

I am not even sure if that's a correct conversion.
But if it is then this "barrier" is extremly small and 99.999% of the beam should make it "through". It's actually not tunneling through anything since the beam energy is higher than the barrier.

But the answer is that ##T = 1.1 \cdot 10^{-54}##.

So I'm comletely lost, because barriers are usually expressed in eV and google keeps telling me you cant convert eV to Volts but it's perfectly fine to convert volts to eV which makes even less sense.

I have the equations to calculate transmission probability, my problem is that the barrier is given in Volts not electron volts.

$$200V = e \cdot 200 eV = 3.2 \cdot 10^{-17} eV$$
This is not correct. If a charge ##q## moves through a potential difference ##\Delta V##, what is the change in the particle's potential energy, ##\Delta U##?

If the charge ##q## is equal to ##e## and the potential difference is 200 Volts, what is ##\Delta U## in Joules? What is ##\Delta U## in eV?

If charge q = e, moves through 200V, the energy given the electron is:

$$\Delta U = qV = 200 V * e = 3.2 \cdot 10^{-17} J$$

I have no clue how to convert that into eV. I havnt done this kind of physics since middleschool, there was no introduction to any of this and next week we have to do statistical quantum mechanics. I need simple equations or pure explanations. Not rethorical questions please :(

But you already described it correctly! If an electron moves in an electrostatic field between two points with the potential difference of 200 V it gains (or looses depending on the direction of motion) 200 eV of kinetic energy.

If charge q = e, moves through 200V, the energy given the electron is:

$$\Delta U = qV = 200 V * e = 3.2 \cdot 10^{-17} J$$
Good.

Please show the formula that you are using to calculate the probability and the numbers (with units) that you are using for all the quantities that appear in the formula.

If charge q = e, moves through 200V, the energy given the electron is:

$$\Delta U = qV = 200 V * e = 3.2 \cdot 10^{-17} J$$

I have no clue how to convert that into eV. I havnt done this kind of physics since middleschool, there was no introduction to any of this and next week we have to do statistical quantum mechanics. I need simple equations or pure explanations. Not rethorical questions please :(
An electron-volt and a joule are both units of energy. You're asking how to do a unit conversion.

Instead of plugging in a value for the elementary charge, the idea is to subsume the factor of ##e## into the unit. That is, for an electron or proton, you have
$$\lvert \Delta U \rvert = e \lvert \Delta V \rvert = e(200~{\rm V}) = 200~{\rm eV}.$$ To convert between ##\Delta U## and ##\Delta V##, it's often just adding or removing an "e" from the unit. So that's why it may seem like volts and electron-volts are interchanged willy-nilly, but that's not really happening.

Note that
$$1~{\rm eV} = e(1~{\rm V}) = 1.60\times 10^{-19}~\rm J$$ so the conversion factor between eV and J is numerically just the value of the elementary charge in coulombs.

Can I add a few thoughts too...

The electron-volt is (a very small) unit of energy and is your friend!

Energy and voltage (potential difference, p.d.) are different types of quantity. But they are related (by E = qV).

If an electron passes though a p.d. of 1volt, it gains (or loses) an amount of potential energy; this amount of energy is called an electron-volt (eV).

For example, suppose you have a p.d. of 12V applied across a lamp. Each electron passing from the negative terminal to the positive terminal will convert 12eV of electrical energy to heat/light energy. (In fact this would take rather a long time as the electrons move very slowly in wire – but there are lots of electrons moving together!)

Note the important conversion: 1eV = 1.60x10⁻¹⁹J

In the Post #1 question, the barrier is produced by a potential difference of 200V. Classically, that means an electron would need at least 200eV to cross the barrier. But the electron's (kinetic) energy is too small - only 50eV, just 1/4 of the barrier 'height'.

vela said:
An electron-volt and a joule are both units of energy. You're asking how to do a unit conversion.
However, "V" is not a unit of energy. It looks like people are assuming that the "50 eV beam" consists of electrons, or of some particle whose |charge| is equal to e. Technically, this was not stated and cannot be assumed, so the OP's confusion seems to be a legitimate one.

That being said, it is probably necessary to assume a 1e charge on the beam particles, whatever they are, or else the problem cannot be solved!

Redbelly98 said:
However, "V" is not a unit of energy. It looks like people are assuming that the "50 eV beam" consists of electrons, or of some particle whose |charge| is equal to e. Technically, this was not stated and cannot be assumed, so the OP's confusion seems to be a legitimate one.
I was responding to the OP's question about converting ##3.2\times 10^{-19}~\rm J## to electron-volts, which is just a unit conversion. The earlier posts had already addressed how to translate the 200-V barrier into the equivalent potential energy barrier. Presumably, the OP knows the charge of the particle involved.

vela said:
I was responding to the OP's question about converting to electron-volts, which is just a unit conversion.
Argh, sorry. I missed that after seeing the "3.2e-17 eV" referred to in Post #1.

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