Question
Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei $\left( ^4\textrm{He} \right)$ from gold-197 nuclei $\left( ^{197}\textrm{Au} \right)$. The energy of the incoming helium nucleus was $8.00 \times 10^{-13} \textrm{ J}$, and the masses of the helium and gold nuclei were $6.68 \times 10^{-27} \textrm{ kg}$ and $3.29 \times 10^{-25} \textrm{ kg}$, respectively (note that their mass ratio is 4 to 197). (a) If a helium nucleus scatters to an angle of $120^\circ$ during an elastic collision with a gold nucleus, calculate the helium nucleus’s final speed and the final velocity (magnitude and direction) of the gold nucleus. (b) What is the final kinetic energy of the helium nucleus?

a) $v_{He}' = 1.49 \times 10^7 \textrm{ m/s}$

$v_{Au}' = 6.16 \times 10^5 \textrm{ m/s, } 29.3^\circ \textrm{ below positive x-axis}$

b) $7.38 \times 10^{-13} \textrm{ J}$

Solution Video

# OpenStax College Physics Solution, Chapter 8, Problem 49 (Problems & Exercises) (17:52)

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This is College Physics Answers with Shaun Dychko. Ernest Rutherford showed that the nucleus of a gold atom is very dense because he showed that helium nuclei which are incident on the gold nucleus will bounce off and go backwards which suggests that this nucleus of gold must be really heavy in order for it to make the helium bounce all the way back. Well, so the scenario we have here is the helium nucleus is incident with an energy of 8.00 times ten to the minus thirteen joules and we're told that it bounces back at an angle of 120 degrees and our job is to figure out what is the angle of recoil of the gold nucleus and its speed and also to figure out the speed of the helium's nucleus recoil. Now we're given the masses of the helium and the gold nuclei and we have three things that we want to find. So what that means is that we will probably have to use three equations to figure it out. Let's put this arrow back here. Okay. So the first equation we get by considering the x direction, conservation of momentum. So initially, we have all of our momentum is included in the helium nucleus and we multiply the mass of the helium nucleus by its initial velocity. We'll figure out what that velocity is a little bit later using this kinetic energy that we're given. After the collision, we're going to have the mass of the helium times its speed, multiplied by cosine of 180 minus this theta he that's given. I like to use the reference angle in my calculations, so I'm using this angle in here. This is 180 minus theta he and because I am doing that I can put a negative sign explicitly in the equation. You could instead have this be positive and then go cosine of theta he or cosine of 120 in other words, that would be fine, but my preference is to make the angle a reference angle so some number that's between zero and 90, and then put positives or negatives in the equation explicitly. So, and then we'll add to that this positive x component of the momentum of the gold nucleus. So that's mass of the gold nucleus multiplied by its speed after the collision, multiplied by cosine of theta au. Okay. Well, there's not much more we can do with that because there are three things we don't know. We don't know this angle here we don't know this speed and we don't know this speed. So we need to consider the y direction to find another equation. So initially there is no momentum at all in the y direction because this helium nucleus is incident straight along the x axis we'll say. Then after the collision the total y component of the momenta must also be zero. So we have mass of the helium nucleus times its speed, times sine of 180 minus theta he and then minus, because this gold is going downwards and to the right and so its y component of its momentum is going to be negative so put a minus there explicitly in the equation, and multiply that gold mass by its speed multiplied by sine of its angle there, theta au. Okay. So that's two equations but we have three unknowns so that means we need a third equation and that comes from conservation of kinetic energy because we are told this is an elastic collision. So we have the total kinetic energy initially, which is one half mass of the helium times its speed squared before the collision, equals one half mass of the helium times its speed after the collision, squared plus one half mass of the gold times the speed of the gold after collision, squared. So, we're going to -- well, we have to look at these equations first of all and think about what is our strategy for solving for the unknowns. Now, one thing I'd like to look at is these angles here because angles are really tricky to solve for because they're inside a trigonometric function. So this theta au, we can't really get at it because it’s contained within this sine function, and likewise for this equation. In the x direction we have theta au is contained inside the cos trigonometric function. So the first thing I like to get rid of are these angles. So our goal is to -- well, next thing to notice is that the factors in front of these trig functions are the same. We have m au v au prime in front of the cos function and then we have m au v au prime in front of the sine function. So that is a clue that if we were to square this factor and add it to this factor squared, we'd have a common factor m au v au all squared that we could factor out  and then we'd have sine squared plus cos squared. Sine squared plus cos squared makes the number one. That's a nice trigonometric identity, that's the Pythagorean identity. So we're going to do some rearrangement here in order to make this possible. So we're going to rearrange things and then we're going to square our rearranged y direction and square our rearranged x direction formulae, and then we'll add them together and end up making the sine and cosine each being squared, added together and that angle will disappear, which is good. Okay. So version b of equation one, we'll isolate this m au v au prime cos theta au term on one side, and so that equals m he v he plus mass of the helium times speed of the helium after collision, times cos 180 minus theta he. That's because I've just moved this term to the left side and then switch the sides around. That's all that happened there and that brings this term isolated on one side of the equation, which is nice because the next step is we're going to square it. Then likewise for the y direction, we're going to isolate the sine theta au term on one side by adding it to both sides or moving it to the left so it becomes positive and then equals the other stuff that was left over there. Then we'll square equation one b, so this is equation one b squared now. So that's m au squared v au prime squared times cos squared theta au, so there's nothing fancy about the left side there. We just squared each of those factors. On the right side it becomes a little bit more messy because we have a binomial and so we'll first of all factor out the common factor, mass of the helium and we'll square that, and then we'll square the v he plus v he prime cos 180 minus theta he. We'll square that binomial and so it's going to be the first term squared plus two times the first term times the second term, then plus the second term squared. Okay. So now, we squared equation two b as well which is straightforward because there is no binomials to consider there, we just square each of the factors. Then we're going to add together the equation one b squared to equation two b squared. So we're adding this equation to this equation and that is shown on the next line. So we have m au v au prime squared as a common factor among these two terms here and here so we can factor it out leaving us with cos theta au squared plus sine squared theta au. This is going to become the number one and so mission accomplished, we got rid of the variable theta au. It doesn't appear anywhere else here. Then on the right hand side we have a common factor m he squared and so we factor that out here and then add everything else together. So that's v he squared plus two v he prime cos 180 minus theta he plus this term and now this v prime helium velocity after collision squared is a common factor between this and this that remains, and so we'll factor that out. We have cos squared 180 minus theta he plus sine squared 180 minus theta he which is another nice coincidence. This also becomes the number one and this becomes the number one. So what we're left with is this line here. We'll call that equation four. So up to this line after we apply the trigonometric identities to cos squared and sine squared here. Well that's a lot of work but still not enough as it turns out because we have two unknowns remaining, the velocity of the helium after collision and the velocity of the gold after the collision. So, let's consider our kinetic energy formula. We're going to rearrange it to solve for v au prime squared because if we can do that we can substitute for it here in equation four. So we rearrange this kinetic energy formula to solve for v au prime squared and multiply everything by two and that gets rid of the fraction there. Then take this term to the left side so it becomes minus and then divide everything by mass of gold and you solve for this. So the velocity of the gold after collision squared is going to be mass of the helium times the initial velocity of the helium squared minus mass of the helium times the after collision velocity of the helium squared, all divided by mass of the gold nucleus. So we'll substitute that in for v au prime squared in equation four in this line now. So you see this is written here in red because it replaced the v au prime squared. Otherwise everything else is the same as in equation four. Then we have some work to do to solve for the velocity of the helium after collision because this is going to end up being a quadratic equation and we can use our quadratic formula to solve it. But we have to change it to the form that we're used to. We need to make it look like a x squared plus b x plus c equals zero. Right now we have things on both sides of the equation and we have brackets and so on. So we have some work to do. So first we'll expand the brackets by multiplying through by the factors outside the brackets and that makes this m au become power of one when it cancels with one of the m au's there. So we have m, mass of the gold times mass of the helium times initial velocity helium squared, minus mass of the gold times mass of the helium times after collision velocity of the helium squared equals m he squared multiplied by each of the terms in there. Then we collect everything on the left side of the equation. Well what I actually did is I moved everything to the right side and then switch the sides around. So we want to collect the like terms together and so we have the v he prime squared terms, there's going to be two of them, there's this one and then there's going to be this one which is going to become positive when it moves to the right side. So we have v he prime squared times mass of the helium squared plus mass of the gold times mass of the helium, plus two times mass of the helium squared times velocity of the helium initially times velocity of the helium after collision. So this is the v he prime to the power of one term. I didn't do anything to that, that's just copied. Then we have mass of the helium times velocity of the helium squared, multiplied by mass of the helium minus mass of the mass of the gold. That comes from combining this term with this term and this term becomes minus when it goes to the right side. These two terms have a common factor of m he times v he squared that I've got factored out, leaving m he to the power of one from this term and leaving m au behind from that term. All that equals zero. Now we have to plug in some numbers there to figure out what our coefficients are in order to make it look like this and to use our quadratic formula. So we need to know what the speed is first of all of the helium nucleus before the collision, and so we know that its kinetic energy we're told is eight times ten to the minus thirteen joules. So we'll solve for v he then by saying that's the square root of two times that divided by the mass of the helium, 6.68 times ten to the minus twenty-seven kilograms. This gives 1.54765 times ten to the seven meters per second. So, now we find all of our coefficients. So we plug in the mass of the helium, squared, add that to the mass of the gold nucleus times the mass of the helium nucleus. That's going to give this coefficient here, and then we have two times mass of the helium nucleus squared times the initial velocity of the helium nucleus. That gives us coefficient here and then we add to that the mass of the helium times the initial speed of the helium squared, times the difference between the helium and the gold nuclei masses. That gives this coefficient here of negative 5.15714 times ten to the minus thirty-seven, and all that equals zero. So then we plug into the quadratic formula to figure out what the velocity of the helium nucleus is after collision. All of this works out to 1.49 times ten to the seven meters per second. Then, we need to figure out the velocity of the gold nucleus after the collision. So we look at equation three b which is here, and we can take the square root of both sides to solve for v au prime. So that's what I did here, and so we've plugged in mass of the helium times initial speed of the helium squared, minus mass of the helium times the final velocity of the helium squared, all divided by the mass of gold nucleus. We get 6.16 times ten to the five meters per second. Then we need to figure out the direction of the gold nucleus and so we look at equation two b and one b and we divide them and we can get a -- this will end up being tangent of theta au because these factors will cancel and we'll have sine theta au divided by cos theta au which is tangent theta au. I chose to do it this way because we'll end up having tangent of this angle equals some stuff containing only one of the numbers that we calculated. I wanted to minimize the number of quantities that we calculated in our equation, so doing it this way results in only using the velocity of the helium after collision. So we have this is going to be our numerator and we're going to divide it by this denominator and we can see there's a common factor m he everywhere so that's going to disappear from down here. So, that worked out to this. So we have velocity of the helium after collision times sine 180 minus theta he, all divided by v he plus v he prime times cos 180 minus theta he. So the angle is going to be the inverse tangent of the velocity of the helium after collision times sine of 60, divided by initial velocity of the helium plus final velocity of the helium times cos of 60. This ends up being 29.3 degrees. So the velocity of the gold nucleus after collision is 6.16 times ten to the five meters per second, 29.3 degrees below the positive x axis. There! If you got this far, congratulate yourself because that was a lot of work to follow. If you got the answer correct on your own then awesome! Because you should be trying these problems before you look at the solutions of course. Now the final kinetic energy of the helium nucleus is going to be one half times its mass times its final velocity squared and that works out to 7.38 times ten to the minus thirteen joules.