Video Transcript
The Bohr radius is a physical
constant that is equal to the distance between the nucleus and the electron of a
hydrogen atom in the ground state. Its value is given by the formula
𝑎 naught is equal to four 𝜋𝜀 naught ℎ bar squared divided by 𝑚 subscript e 𝑞
subscript e squared. Calculate the value of the Bohr
radius. 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 in scientific
notation to two decimal places.
So we’ve been given this formula
here for the Bohr radius 𝑎 naught. And our task is to use this formula
to calculate the value of this Bohr radius. We can see that in the second half
of the question text we’re given the values of all the different quantities that
appear on the right-hand side of the formula.
In order 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 were given. This here is the formula we were
given in the question. In this formula, we were told that
the permittivity of free space, that’s the quantity 𝜀 naught, is equal to 8.85
times 10 to the negative 12 farads per meter. We’re also told that the reduced
Planck constant, so that’s ℎ bar, has a value of 1.05 times 10 to the negative 34
joule-seconds. As a quick aside, we might recall
that this reduced Planck constant ℎ bar is simply equal to the regular Planck
constant with a symbol ℎ divided by two 𝜋.
For the rest mass of an electron 𝑚
subscript e, we’re given a value of 9.11 times 10 to the negative 31 kilograms. And for the charge of an electron
𝑞 subscript e, we’ve got a value of 1.60 times 10 to the negative 19 coulombs. It’s worth briefly mentioning that
this value here is actually the magnitude of the electron charge. The electron is a negatively
charged particle, which means that its actual charge would have a negative sign out
front. In this formula for the Bohr
radius, we’re simply concerned with the magnitude of this charge. In fact, since 𝑞 subscript e
appears to the power of two in this formula, then even if we did include a negative
sign in the electron charge, when we square it, we’d have a negative times a
negative. So the negative signs would just
cancel out.
We know that this quantity 𝑎
naught that we’re calculating is this constant known as the Bohr radius. The question told us that this Bohr
radius is equal to the distance between the nucleus and the electron in the ground
state of a hydrogen atom. Recall that a hydrogen atom is an
atom that has just one proton as its nucleus and just one electron orbiting
this. Let’s also recall that in the Bohr
model of the atom, there’s a positively charged nucleus at the center, which we’ve
represented with this red circle here. And then negatively charged
electrons orbit this nucleus on circular orbits of particular-defined radii that
we’ve shown some of as these black circles on our diagram.
The particular one of these circles
that are given electron orbits on depends on the energy level of the electron. An electron with an energy level of
one, which is the lowest possible energy level, is on the orbit that’s closest to
the nucleus. The next circular orbital
corresponds to an energy level of two. And the next one is three, and so
on for larger values of the energy level.
Now we’re thinking here about a
hydrogen atom which we know has just one electron. And we know that the Bohr radius is
the distance between the nucleus and the electron when the hydrogen atom is in its
ground state. An atom is in its ground state when
all of its electrons are in the lowest possible energy levels that they can be
in. For hydrogen then, with just a
single electron, in the ground state this electron must have an energy level of
one. And so it’ll be on the innermost
circular orbit nearest to the nucleus. In our sketch, we’ve shown this
with this blue circle here representing the electron.
So we’ve now got a sketch of
hydrogen in its ground state according to the Bohr model of the atom. We know that the Bohr radius is the
distance between the nucleus and the electron in this ground state of hydrogen. So in our sketch, that’s the
distance out from this nucleus or proton to the closest-in circular orbit, which we
know is the orbit that the electron will be on in the ground state.
Now that we’ve recapped what it is
that this Bohr radius we’re calculating actually represents, let’s go ahead and work
out its value. To do this, all that we need to do
is to take these values that we’re given and substitute them into this formula and
then evaluate the right-hand side of the expression. Let’s clear ourselves some space on
the board so we can make a start on this.
Substituting these values into this
formula, we get this expression here for the Bohr radius 𝑎 naught. So in the numerator, that’s four 𝜋
multiplied by the value of 𝜀 naught, that’s the permittivity of free space,
multiplied by the square of the reduced Planck constant ℎ bar. This then all gets divided by the
rest mass of an electron 𝑚 subscript e and the square of the charge of an electron
𝑞 subscript e. Notice that all of the units on the
right-hand side either are SI base units or, in the case of farads, joules, and
coulombs, can be expressed purely in terms of SI base units. That means that when we use these
values to calculate the Bohr radius 𝑎 naught, it will be given in the SI base unit
for distance, which is meters.
With this in mind, let’s now tidy
up this expression. In this second expression, rather
than writing out all the different units on the right-hand side, we’ve simply
written the units of meters that we know that our result for 𝑎 naught will
have. We’ve also separated out all of
these values from all of these powers of 10. This means we can then work out
these two parts of the expression separately before multiplying them together.
Let’s start by considering the
powers of 10. For this, there are two identities
that we’re going to find helpful. The first is that if we have some
value 𝑐 raised to the power of 𝑥 multiplied by that same value 𝑐 to the power of
𝑦, then this is equal to 𝑐 to the power of 𝑥 plus 𝑦. The second identity is that 𝑐 to
the 𝑥 divided by 𝑐 to the 𝑦 is equal to 𝑐 to the power of 𝑥 minus 𝑦.
Now in this term in our expression,
the numerator reads 10 to the negative 12 multiplied by the square of 10 to the
negative 34. We can write out the square of 10
to the negative 34 as 10 to the negative 34 times 10 to the negative 34. Then, by comparing against this
identity here, we see that we can rewrite the numerator as 10 to the power of
negative 12 minus 34 minus 34, which works out as 10 to the power of negative
80.
We can now do the same thing in the
denominator. Here we’ve got 10 to the negative
31 multiplied by the square of 10 to the negative 19. This can be rewritten as 10 to the
negative 31 minus 19 minus 19, which comes out as 10 to the power of 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 this fraction as 10 to the power of negative 80
minus negative 69. These two minus signs here end up
negating each other, giving us 10 to the power of negative 80 plus 69, which works
out as 10 to the power of negative 11.
Now that we’ve simplified this
second term, 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. We can type this into a calculator
to get a result of 122.6116 and so on with further decimal places. Then in the denominator, we have
9.11 multiplied by the square of 1.60. This works out as exactly
23.3216. Dividing this numerator by this
denominator, we get a result of 5.2574 with further decimal places.
Looking at these two parts that we
found together, we can see that we’ve now got a value for the Bohr radius 𝑎
naught. We found that it’s equal to 5.2574
et cetera times 10 to the negative 11 meters.
The last thing to recall is that
the question wanted our answer in scientific notation to two decimal places. The value that we found for the
Bohr radius is already in scientific notation, so we just need to round it to two
decimal places. To do this, we need to check the
third decimal place, which we see is equal to seven. Since this is greater than or equal
to five, this means that the value in the second decimal place gets rounded up.
And so our answer is that in
scientific notation, to two decimal places, the Bohr radius has a value of 5.26
times 10 to the negative 11 meters.
