### Video Transcript

Use the formula π subscript π is
equal to four ππ naught β bar squared π squared divided by π subscript π π
subscript π squared, where π subscript 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 subscript π is the mass of the electron, and
π subscript π is the charge of the electron to calculate the orbital radius of an
electron that is in energy level π 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, weβre given a formula here
which tells us how to calculate the orbital radius π subscript π of an electron in
energy level π of a hydrogen atom. Weβre being asked to use this
formula to find the value of π subscript π for an electron thatβs in energy level
π equals four. In the second half of the question
text, weβre given the values of all the different constants on the right-hand side
of the equation. To answer this question, weβll need
to clear ourselves some space on the board. But as we do this, letβs make a
note of all these values that weβre given.

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

The other values that weβre given
are for the mass and the charge of an electron. Weβre told that the electron rest
mass π subscript π is equal to 9.11 times 10 to the negative 31 kilograms. And the electron charge π
subscript π is 1.60 times 10 to the negative 19 coulombs. The last thing to recall from the
question is that weβre asked to find the value of π subscript π for the case that
π 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,
thereβs a nucleus at the center, which weβve 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
electronβs value of π, where π is the energy level of the electron and is also
referred to as the electronβs principal quantum number. π equals one corresponds to the
innermost circular orbit. So, thatβs the orbit nearest the
nucleus. π equals two corresponds to the
next one out. And then weβve got π equals three,
π equals four, and so on for larger values of π.

Now, this sketch that weβve drawn
here certainly isnβt drawn to scale. However, it does at least give us
some sense that for larger values of the energy level π, the size of the orbital
radius of the electron π subscript π is bigger. This equation from the Bohr model
is describing this relationship between the energy level π and the radius of the
orbit π subscript π. If we rewrite the equation like
this, pulling the π 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 π subscript π is proportional to π 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 π subscript naught. Then, using the Bohr radius, we can
say that the orbital radius π subscript π is equal to the Bohr radius π subscript
naught multiplied by π squared.

Now, all of this discussion has
actually been somewhat of a detour because in the question weβre given all of these
values that can go directly into this equation here. Recall that in the question, weβre
told that weβre 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 weβve
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 weβre told in this case is in the energy level π equals
four.

So, letβs add an electron to the
fourth circle out from the nucleus in our sketch. Thatβs this small blue circle that
weβve added here. Now that weβve gained some
understanding of what it is that this equation is describing, letβs 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
weβre calculating is π subscript four. So thatβs the value of π subscript
π when π is equal to four. Substituting in those values gives
us this expression here. So, in the numerator thatβs four π
multiplied by the value of π naught, the permittivity of free space, multiplied by
the square of β bar, the reduced Planck constant, multiplied by the square of our
value of π, the energy level of the electron. All of this then gets divided by
the rest mass of the electron, π subscript π, and the square of the charge of the
electron, π subscript π.

Now, letβs consider the units in
this expression. The value of π 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 itβs 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 β 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. Itβs equal to the kilogram meter
squared per second squared. The last remaining quantity in the
numerator is the energy level π. And this is a dimensionless
quantity, which means that it has no units at all.

In the denominator, weβve got the
rest mass of the electron in units of kilograms. Thatβs 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. Weβve 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
weβll calculate will be in the SI base unit for distance, which is units of
meters. With this in mind, letβs tidy up
this expression a bit.

In this second expression, rather
than writing out all of the units on the right-hand side, weβve simply written the
units of meters that we know this orbital radius will have. Weβve 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.

Letβs begin by considering the
powers of 10. Thereβs two identities that weβre
going to find helpful. The first one is that π to the
power of π₯ multiplied by π to the power of π¦ is equal to π to the power of π₯
plus π¦. The second identity is that π to
the π₯ divided by π to the π¦ is equal to π to the power of π₯ minus π¦.

In this term in our expression, in
the numerator weβve 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, letβs 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, weβve now got an expression for
the orbital radius π subscript four in units of meters. However, rather than units of
meters, because orbital radii are so small, itβs 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 π subscript
four then, letβs 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 weβve 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 π 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 π 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 π equals
four in hydrogen is equal to 0.84 nanometers.