Past JEE Questions: Scattering

Transcript

0:00Introduction

Hello everyone! Fire alpha particles at a very thin gold foil, and most of them go straight through. But about one in eight thousand turns back, through more than ninety degrees. How can a thin foil throw a particle back? That surprise gave us the nucleus, and it comes up often in JEE papers. Today, we crack four real past questions with two formulas and one clear idea. Let's begin.

0:29Formula one: the distance of closest approach

So, formula one. Look at the bottom path in the figure: an alpha particle fired head-on at a heavy nucleus, which we take as fixed; forget the upper two paths for now. Alpha, the Greek letter like a little fish, has charge two e, and the nucleus, the red dot, has charge plus Z e, with Z its atomic number. The repulsion slows the alpha particle down. So at the closest point, what is its speed? Yes, exactly, the speed is zero for one instant, so all its kinetic energy, capital K, has become potential energy. K equals one over four pi epsilon naught, times two e times Z e, over r naught; the brackets keep each charge together. Epsilon naught, the curly e with a small zero, is the permittivity of free space, and r naught is the distance of closest approach. With me? Now rearrange this for r naught. r naught equals one over four pi epsilon naught, times two Z e squared, over K; the small two up top means squared. Okay? Now the shortcut that saves time. e squared over four pi epsilon naught equals one point four four M e V femtometre, where M e V is the mega electron volt, and f m is the femtometre, the unit of length for nuclei. Right? So, for an alpha particle, two times one point four four gives two point eight eight. r naught equals two point eight eight times Z over K, with K in M e V, shown in brackets, and the answer comes directly in femtometres. Clear?

2:00JEE Main 2020: the graph of N against θ

Now formula two, with a question from JEE Main twenty twenty: which graph shows N, the number of scattered alpha particles, against theta, the scattering angle? Theta is the Greek letter like an o with a line across it. N is proportional to one over sine to the power four of theta over two, for the same foil and the same beam; the stretched alpha sign means proportional to. Okay? So, when theta is small, the sine of theta over two is tiny, and we divide by the fourth power of that tiny number. So is N large or small at small angles? It is huge, that is correct, because a tiny number to the power four is tinier still. So most scattered particles are found at small angles. Got it? Now look at one hundred and eighty degrees; the small circle on screen means degrees. As theta grows, N keeps falling, so at one hundred and eighty degrees N becomes zero, right? Be careful, because this is exactly the trap in this question. At one hundred and eighty degrees, theta over two is ninety degrees, and sine ninety is one. So N is at its smallest, but not zero: some particles really do bounce straight back. With me? So the correct graph is very high at small angles, falls steeply, and stays small but non-zero all the way to one hundred and eighty degrees. Clear?

3:23JEE Main 2024: Rutherford's model or Thomson's?

Now the one clear idea, from JEE Main twenty twenty four. Statement one: in Rutherford's model, the positive charge is in the nucleus, and the electrons revolve round it. Statement two: the plum pudding model is a special case of Rutherford's model. Is statement two true or false? Start with Rutherford: the positive charge sits in a tiny nucleus, and the electrons revolve round it. That is exactly statement one, so it is true. Okay? Now Thomson's plum pudding model. Here the positive charge is spread over the whole atom, with the electrons embedded in it, like seeds in a watermelon. Right? So it predicts only very small deflections, because the spread-out charge never pushes hard enough. But in the experiment, a few alpha particles turned back. Got it? So the plum pudding model has no nucleus at all; it is a different, older model that the experiment ruled out, not a special case. Statement one is true, and statement two is false, and that is the answer.

4:30Example 1: closest approach of a 7.7 MeV alpha to gold

Now our first example, from JEE Main twenty twenty six. An alpha particle of energy seven point seven M e V is fired at a thin gold foil, with Z equal to seventy nine. Find the closest distance it can reach from a gold nucleus. Pause the video and try it first. Done? Okay, let's check. e squared over four pi epsilon naught is one point four four. So r naught is one point four four times seventy nine, over seven point seven, right? That misses a factor of two, so be careful. The alpha particle has charge two e, so we need two point eight eight. And the closest distance comes in a head-on collision, so r naught equals two point eight eight times Z over K, in femtometres. Okay? Now substitute. Two point eight eight times seventy nine, the cross means times, is two hundred and twenty seven point five two. Divide that by seven point seven, and we get about twenty nine point five femtometres; the wavy equals sign means approximately. With me? Finally, the answer in metres. One femtometre is ten to the power minus fifteen metre. So twenty nine point five femtometres is two point nine five times ten to the power minus fourteen metre.

5:51Example 2: r0 is proportional to 1/m (AIEEE 2006)

Clear? Now a slightly harder one, from the old A I triple E paper of two thousand six. An alpha particle of energy half m v squared, with m its mass and v its speed, bombards a heavy nucleus of charge Z e. Its distance of closest approach is proportional to which of these: v squared, one over m, one over v to the power four, or one over Z e? Pause the video and try it first. Done? Okay, let's check. In formula one, put K equal to half m v squared. So r naught equals one over four pi epsilon naught, times two Z e squared, over half m v squared. Okay? Now, four pi epsilon naught, the two, the half and e squared are all constants. So r naught is proportional to Z over m v squared. Clear? Here is where many students slip. The speed is already squared inside the energy, and then the energy sits in the bottom. So r naught goes as one over v to the power four, right? No, the speed gets squared only once, so be careful here. r naught is proportional to one over v squared, so both v squared and one over v to the power four are wrong. Got it? Now the last option, one over Z e. Z sits on top in our formula, so r naught is proportional to Z, and that option is out as well. With me? That leaves one over m, and it is correct. At the same speed, a heavier particle carries more energy, so it gets closer, and r naught is smaller. Nice, right?

7:28Quick revision: r0 = 2.88 Z/K fm, N vs theta, Thomson model

So, quick revision. Head-on, r naught is two point eight eight times Z over K femtometres, with K in mega electron volts. r naught goes as Z over m v squared, and the speed is squared once, not twice. N goes as one over sine to the power four of theta over two: huge at small angles, never zero at one hundred and eighty degrees. And Thomson's model has no nucleus, so it is not a special case of Rutherford's. Now go and try the practice problems in this chapter. All the best!