# Video: Calculating the Orbital Radius of an Electron in the Hydrogen Atom

Use the formula π_π = 4ππββΒ²πΒ²/π_e π_eΒ², where π is the orbital radius of an electron in energy level π of a hydrogen atom, πβ is the permittivity of free space, β is the reduced Planck constant, π_e is the mass of the electron, and π_e is the charge of the electron, to calculate the orbital radius of an electron that is in energy level π = 2 of a hydrogen atom. Use a value of 8.85 Γ 10β»ΒΉΒ² F/m for the permittivity of free space, 1.05 Γ 10β»Β³β΄ Jβs for the reduced Planck constant, 9.11 Γ 10β»Β³ΒΉ kg for the rest mass of an electron, and β1.60 Γ 10β»ΒΉβΉ C for the charge of an electron. Give your answer to 3 significant figures.

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### Video Transcript

Use the formula π π equals four ππ naught β bar squared π squared over π e π e squared, where π is the orbital radius of an electron in energy level π of a hydrogen atom, π naught is the permittivity of free space, β bar is the reduced Planck constant, π e is the mass of the electron, and π e is the charge of the electron, to calculate the orbital radius of an electron that is in energy level π equals two of a hydrogen atom. Use a value of 8.85 times 10 to the negative 12 farads per meter for the permittivity of free space, 1.05 times 10 to the negative 34 joule seconds for the reduced Planck constant, 9.11 times 10 to the negative 31 kilograms for the rest mass of an electron, and negative 1.60 times 10 to the negative 19 coulombs for the charge of an electron. Give your answer to three significant figures.

Okay, so this seems like a pretty long question. But actually, all weβre being asked to do is this bit: Calculate the orbital radius of an electron that is in energy level π equals two of a hydrogen atom. The rest of the question just tells us how we can do this. So weβre told we can use this formula. And this part of the question defines what all the quantities in this formula are. And this last part of the question tells us the values of the constants in the equation.

We can recall that this formula which weβve been given is derived from the Bohr model of the atom, which describes atoms as consisting of a positively charged nucleus with electrons making circular orbits around it. Now, the Bohr model has some limitations, but itβs still pretty accurate when describing systems with one electron such as the hydrogen atom in this question.

Now, in this question, weβre told that the electron occupies energy level π equals two. We can recall that π is the principal quantum number of an electron in an atom. π takes whole number values, which describe the energy level that an electron has. The lowest value that π can take is one, which would describe an electron in the lowest possible energy state of an atom. In the Bohr model, this would describe an electron in the innermost orbital around the nucleus.

However, in this question, weβre told that our electron is in energy level π equals two, which means that the electron occupies the next orbital out. The orbital radius of an electron is simply the radius of the circular path that it follows around the nucleus. And as weβve been told in the question, we can calculate the orbital radius of an electron using this equation.

One interesting thing to note about this equation is that it only actually contains two variables, the orbital radius and the principal quantum number. This means that according to the Bohr model, the orbital radius of an electron is proportional to the square of its principal quantum number. Now, we want to calculate the orbital radius of an electron in energy level π equals two. In other words, weβre looking to find π two. To find this, we simply substitute two in place of π in this equation, which gives us four ππ naught β bar squared two squared over π e π e squared, where weβve been told π naught is the permittivity of free space. β bar is the reduced Planck constant. π e is the mass of an electron. And π e is the charge of an electron.

Since weβre told the values of all of these quantities in the question, all we need to do now is substitute these values in and calculate the answer, which gives us this expression. And if we plug all of this into our calculators, it gives us an answer of 2.10 times 10 to the power of negative 10 meters, which is equivalent to 0.210 nanometers. And this is the final answer to our question. The orbital radius of an electron in energy level π equals two of a hydrogen atom is 0.210 nanometers.