### Video Transcript

An absorption line normally at 552
nanometers appears at 585 nanometers in the spectrum of light coming from a
galaxy. How fast is the galaxy moving away
from Earth? Give your answer in kilometers per
second and in scientific notation to three significant figures.

Okay, so in this question, we’re
looking at an absorption line or, in other words, a line on an absorption spectrum,
which we have been told normally appears at 552 nanometers. Now, when we say normally, what we
mean is that in an absorption spectrum taken on Earth, this particular absorption
line appears at 552 nanometers. However, in the absorption spectrum
taken from the light coming from a galaxy, this same absorption line appears at 585
nanometers. In other words then, we can see
that the wavelength of light coming from this galaxy has increased relative to the
wavelength of light that we would see on Earth.

Now this increase in wavelength of
light coming from the galaxy is known as red shift because the light is shifted to
the red end of the spectrum. And this red shift occurs because
the source of the light in question is moving away from the observer. In other words, when the absorption
line that we said was normally at 552 nanometers was found, the source of that light
was on Earth itself. And therefore was not moving
relative to the observer. However, the galaxy from which
we’re gathering light does seem to be moving away from the observer, which in this
case is a scientist on Earth. And because of that, the light
coming from the galaxy is red shifted.

Now, we’ve been given the original
wavelength of a particular absorption line as well as the wavelength it moves to in
the light coming from this particular galaxy. So based on these numerical values,
we need to work out how fast the galaxy is moving away from Earth. To do this, we need to recall the
equation that tells us that Δ𝜆, the difference in wavelength of the absorption
line in the light from the galaxy and the light from a source that is stationary
relative to the observer, divided by the wavelength of the spectral line, from the
source of light that’s stationary relative to the observer, is equal to the velocity
of the galaxy away from Earth divided by the speed of light.

And at this point, we can see that
we’ve got enough information to answer this question. Because Δ𝜆, the difference in
wavelengths of the absorption line, is simply given as Δ𝜆 is equal to 𝜆 subscript
galaxy, the wavelength of the absorption line from the light from the galaxy, minus
𝜆 subscript rest, which is once again the wavelength of the spectral line in the
light coming from a source that is stationary relative to the observer. And we have values for both 𝜆
galaxy and 𝜆 rest. So we can work out what Δ𝜆 is in
the numerator of this equation. And additionally, we have 𝜆 rest
and we know that the speed of light, 𝑐, is about three times 10 to the power of
eight meters per second.

So we can take this equation and
rearrange it to solve for 𝑣, the velocity of the galaxy relative to the earth. So let’s go about doing that. We can do this by multiplying both
sides of the equation by the speed of light, 𝑐. This way, the speed of light on the
right-hand side will cancel out. And what we’re left with is that
the difference in wavelength between the two spectral lines multiplied by the speed
of light 𝑐 divided by the spectral line’s wavelength, when coming from a source
that’s stationary relative to the observer, is equal to the speed of the galaxy away
from Earth, 𝑣.

So now that we know this, let’s
plug in some values. We can say that the speed of the
galaxy 𝑣 is equal to firstly Δ𝜆, which we know is 𝜆 sub galaxy minus 𝜆 sub rest
or, in other words, 585 nanometers minus 552 nanometers. And then we multiply this
parenthesis by the speed of light, which we know is three times 10 to the power of
eight meters per second. And then we have to divide this
whole thing by 𝜆 sub rest, which once again we know is 552 nanometers.

Now, firstly, when we carry out
this subtraction, we see that it becomes 33 nanometers. And then we can see that the unit
of nanometers in the numerator cancels with the unit of nanometers in the
denominator. Therefore, our final unit is going
to be meters per second. And to find the numerical value,
all we need to do is to find 33 multiplied by three times 10 to the power of eight
divided by 552. Altogether then, we find that the
velocity of the galaxy away from Earth is equal to 0.17935, dot, dot, dot times 10
to the power of eight meters per second. But remember, we need to give our
answer in kilometers per second and in scientific notation and to three significant
figures.

So let’s start by putting our
answer in scientific notation. To do this, we can recall that a
number in scientific notation is written as 𝑎 times 10 to the power of 𝑏. Where 𝑎 is any number between one
and 10, one is included in this range but 10 is not, and 𝑏 is an integer. So based on this information, we
can see that our number over here is not in scientific notation. Because our value of 𝑎 does not
lie between one and 10.

So to remedy this, what we can do
is to think about what this 10 to the power of eight means. 10 to the power of eight simply
means 10 times 10 times 10, and so on and so forth. We’re multiplying this eight
times. And so, what we can do is to
transfer one of these times 10s into our value of 𝑎. What that leaves us with then is
1.7935. That’s our value of 𝑎 now. And we multiply this by 10 to the
power of seven. Because remember, we took this
times 10 and moved it into our value of 𝑎. So there’s one fewer powers of 10
left in this part here. And so, we can say that the value
of the velocity of the galaxy moving away from Earth is 1.7935 times 10 to the power
of seven meters per second.

So we’ve dealt with the scientific
notation part. Now, let’s convert this to
kilometers per second. To do this, we can recall that one
meter is equivalent to one thousandth of a kilometer. And therefore, one meter per second
is equal to one thousandth of a kilometer per second. And hence, to convert from meters
per second to kilometers per second, we need to divide this number by 1000. Or, in other words, we need to
divide it by 10 to the power of three, which is 1000. And so, we can recall that 10 to
the power of seven is equal to 10 times 10 times 10 times — seven times. And 10 to the power of three is 10
times 10 times 10. So three of the powers of 10 are
going to be cancelled from the numerator. And we’re going to be left with 10
to the power four. At which point, we’ve converted to
kilometers per second. So tidying everything up a bit, we
see that our velocity is 1.7935 times 10 to the power of four kilometers per
second. So we’ve now converted to
kilometers per second as well.

The last thing we need to worry
about is rounding our answer to three significant figures. So here’s significant figure number
one, number two, and number three. Now we look at the next significant
figure, which is this value here. It’s a three. And so, because three is less than
five, our third significant figure is going to stay the same. It’s not going to round up. And so, at this point, we’ve found
the answer to our question, to three significant figures as well. We can say that the speed at which
the galaxy’s moving away from Earth is 1.79 times 10 to the power of four kilometers
per second. Given in the correct units of
kilometers per second and in scientific notation and to three significant
figures.