Question Video: Calculating the Orbital Radius of an Electron Using the Bohr Model Physics • 9th Grade

Use the formula 𝑟_(𝑛) = (4𝜋𝜀₀ ℎ bar² 𝑛²)/(𝑚_(𝑒) 𝑞_(𝑒) ^(2))), where 𝑟_(𝑛 ) is the orbital radius of an electron in energy level 𝑛 of a hydrogen atom, 𝜀₀ is the permittivity of free space, ℎ bar is the reduced Planck constant, 𝑚_(𝑒) is the mass of the electron, and 𝑞_(𝑒) is the charge of the electron to calculate the orbital radius of an electron that is in energy level 𝑛 = 4 of a hydrogen atom. Use a value of 8.85 × 10⁻¹² Fm⁻¹ for the permittivity of free space, 1.05 × 10^(−34) J ⋅ s for the reduced Planck constant, 9.11 × 10^(−31) kg for the rest mass of an electron, and 1.60 × 10^(−19) C for the charge of an electron. Give your answer to two decimal places.

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Video Transcript

Use the formula 𝑟 subscript 𝑛 is equal to four 𝜋𝜀 naught ℎ bar squared 𝑛 squared divided by 𝑚 subscript 𝑒 𝑞 subscript 𝑒 squared, where 𝑟 subscript 𝑛 is the orbital radius of an electron in energy level 𝑛 of a hydrogen atom, 𝜀 naught is the permittivity of free space, ℎ bar is the reduced Planck constant, 𝑚 subscript 𝑒 is the mass of the electron, and 𝑞 subscript 𝑒 is the charge of the electron to calculate the orbital radius of an electron that is in energy level 𝑛 equals four 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 1.60 times 10 to the negative 19 coulombs for the charge of an electron. Give your answer to two decimal places.

So, we’re given a formula here which tells us how to calculate the orbital radius 𝑟 subscript 𝑛 of an electron in energy level 𝑛 of a hydrogen atom. We’re being asked to use this formula to find the value of 𝑟 subscript 𝑛 for an electron that’s in energy level 𝑛 equals four. In the second half of the question text, we’re given the values of all the different constants on the right-hand side of the equation. To answer this question, we’ll need to clear ourselves some space on the board. But as we do this, let’s make a note of all these values that we’re given.

This here is the formula from the question. In this equation, we’re told that 𝜀 naught, that’s the permittivity of free space, has a value of 8.85 times 10 to the negative 12 farads per meter. We’re also told to use a value for the reduced Planck constant ℎ bar of 1.05 times 10 to the negative 34 joule seconds. As a quick aside, we might recall that the reduced Planck constant ℎ bar is simply equal to the regular Planck constant ℎ divided by two 𝜋.

The other values that we’re given are for the mass and the charge of an electron. We’re told that the electron rest mass 𝑚 subscript 𝑒 is equal to 9.11 times 10 to the negative 31 kilograms. And the electron charge 𝑞 subscript 𝑒 is 1.60 times 10 to the negative 19 coulombs. The last thing to recall from the question is that we’re asked to find the value of 𝑟 subscript 𝑛 for the case that 𝑛 is equal to four.

Now, this equation that we were given describes the Bohr model of the atom. In the Bohr model of the atom, there’s a nucleus at the center, which we’ve represented with this red circle here. And then, this nucleus is surrounded with these various different circular orbits that an electron can occupy. The particular one of these circular orbits that a particular electron will be found on depends on that electron’s value of 𝑛, where 𝑛 is the energy level of the electron and is also referred to as the electron’s principal quantum number. 𝑛 equals one corresponds to the innermost circular orbit. So, that’s the orbit nearest the nucleus. 𝑛 equals two corresponds to the next one out. And then we’ve got 𝑛 equals three, 𝑛 equals four, and so on for larger values of 𝑛.

Now, this sketch that we’ve drawn here certainly isn’t drawn to scale. However, it does at least give us some sense that for larger values of the energy level 𝑛, the size of the orbital radius of the electron 𝑟 subscript 𝑛 is bigger. This equation from the Bohr model is describing this relationship between the energy level 𝑛 and the radius of the orbit 𝑟 subscript 𝑛. If we rewrite the equation like this, pulling the 𝑛 squared term out to the side, then we can notice that all of these terms inside the parentheses are simply constants. That means then that the orbital radius 𝑟 subscript 𝑛 is proportional to 𝑛 squared.

As an aside, all of these constant terms inside the parentheses often get grouped together as a single constant known as the Bohr radius with a symbol of 𝑎 subscript naught. Then, using the Bohr radius, we can say that the orbital radius 𝑟 subscript 𝑛 is equal to the Bohr radius 𝑎 subscript naught multiplied by 𝑛 squared.

Now, all of this discussion has actually been somewhat of a detour because in the question we’re given all of these values that can go directly into this equation here. Recall that in the question, we’re told that we’re dealing with a hydrogen atom, which is an atom that has just one proton as its nucleus and just one electron orbiting this. So then, if this sketch that we’ve drawn is of a hydrogen atom, then this red circle in the middle that we said was the nucleus is actually just a single proton. And we know that the atom has just a single electron, which we’re told in this case is in the energy level 𝑛 equals four.

So, let’s add an electron to the fourth circle out from the nucleus in our sketch. That’s this small blue circle that we’ve added here. Now that we’ve gained some understanding of what it is that this equation is describing, let’s go ahead and substitute these values into that equation to calculate the value of the orbital radius for this electron.

When we do this, the quantity that we’re calculating is 𝑟 subscript four. So that’s the value of 𝑟 subscript 𝑛 when 𝑛 is equal to four. Substituting in those values gives us this expression here. So, in the numerator that’s four 𝜋 multiplied by the value of 𝜀 naught, the permittivity of free space, multiplied by the square of ℎ bar, the reduced Planck constant, multiplied by the square of our value of 𝑛, the energy level of the electron. All of this then gets divided by the rest mass of the electron, 𝑚 subscript 𝑒, and the square of the charge of the electron, 𝑞 subscript 𝑒.

Now, let’s consider the units in this expression. The value of 𝜀 naught is given in units of farads per meter. The meter is the SI base unit for distance. The farad is an SI-derived unit used to measure capacitance. While it’s not a particularly pretty expression, the farad can be expressed in SI base units as seconds to the four ampere squared meters to the minus two kilograms to the minus one. The reduced Planck constant ℎ bar is given in units of joule seconds. The second is the SI base unit for time. And just as with the farad, the joule can be expressed in SI base units. It’s equal to the kilogram meter squared per second squared. The last remaining quantity in the numerator is the energy level 𝑛. And this is a dimensionless quantity, which means that it has no units at all.

In the denominator, we’ve got the rest mass of the electron in units of kilograms. That’s the SI base unit for mass. And this is multiplied by the square of the electron charge in units of coulombs. The coulomb can be expressed in SI base units as ampere seconds. We’ve seen that all of the units on the right-hand side, either are SI base units or can be expressed in terms of SI base units. This means that the orbital radius we’ll calculate will be in the SI base unit for distance, which is units of meters. With this in mind, let’s tidy up this expression a bit.

In this second expression, rather than writing out all of the units on the right-hand side, we’ve simply written the units of meters that we know this orbital radius will have. We’ve also separated out all of these values from all of these powers of 10. This means we can work out these two parts of the expression separately and then multiply them together.

Let’s begin by considering the powers of 10. There’s two identities that we’re going to find helpful. The first one is that 𝑎 to the power of 𝑥 multiplied by 𝑎 to the power of 𝑦 is equal to 𝑎 to the power of 𝑥 plus 𝑦. The second identity is that 𝑎 to the 𝑥 divided by 𝑎 to the 𝑦 is equal to 𝑎 to the power of 𝑥 minus 𝑦.

In this term in our expression, in the numerator we’ve got 10 to the negative 12 multiplied by the square of 10 to the negative 34. We can rewrite the square of 10 to the negative 34 as 10 to the negative 34 multiplied by 10 to the negative 34. Then, comparing against this identity, we see that we can rewrite the numerator as 10 to the power of negative 12 minus 34 minus 34. That works out as 10 to the power of negative 80.

We can then do the same thing in the denominator, where we have 10 to the negative 31 multiplied by the square of 10 to the negative 19. We can write this out as the product of three powers of 10 and then use this identity to rewrite it as 10 to the negative 31 minus 19 minus 19, which comes out as 10 to the negative 69. So, we now have 10 to the negative 80 divided by 10 to the negative 69.

Comparing against this second identity, we see that we can rewrite our fraction as 10 to the power of negative 80 minus negative 69. These two minus signs here end up negating each other. And so, we have 10 to the negative 80 plus 69, which works out as 10 to the power of negative 11.

Now, let’s turn our attention to the first term in the expression. In the numerator, we have four 𝜋 multiplied by 8.85 multiplied by 1.05 squared multiplied by four squared. We can type this into a calculator to get a result of 1961.7864 and so on with further decimal places. Meanwhile, in the denominator, we have 9.11 multiplied by the square of 1.60. This works out as exactly 23.3216. Then, dividing this numerator by this denominator, we get a value of 84.1189 with further decimal places.

So, we’ve now got an expression for the orbital radius 𝑟 subscript four in units of meters. However, rather than units of meters, because orbital radii are so small, it’s more usual to express them in units of nanometers. We can recall that one nanometer is equal to 10 to the negative nine meters. If we multiply both sides by 10 to the nine, then on the right we have 10 to the negative nine times 10 to the nine, which is simply equal to one. And so we have that one meter is equal to 10 to the nine nanometers.

In our expression for 𝑟 subscript four then, let’s replace these units of meters with units of 10 to the nine nanometers. If we move this 10 to the nine inside this set of parentheses, then we’ve got 10 to the negative 11 multiplied by 10 to the nine, which we can write as 10 to the negative 11 plus nine. And this then works out as 10 to the negative two. So now we have 𝑟 subscript four in units of nanometers.

We can simplify this expression further by recognizing that multiplying by 10 to the negative two is equivalent to moving the decimal point two places to the left. This gives us that 𝑟 subscript four is equal to 0.841189 et cetera nanometers. We can recall that the question wanted our answer to two decimal places. So then, rounding our result, we get our answer that the orbital radius of an electron in the energy level 𝑛 equals four in hydrogen is equal to 0.84 nanometers.