AIEEE and Earlier JEE Main Questions: Bohr Model

Transcript

0:00Introduction

Hello everyone! Let us start with a quick question. In the ground state of hydrogen, the electron goes once round the nucleus in about one point six times ten to the power minus sixteen seconds. Now the electron moves up to the first excited state, where the quantum number n is two instead of one. So n doubles, and the time for one round also doubles, right? Many students say yes. Hold on to that answer, because a JEE Main question from twenty twenty is built on exactly this idea. Today we solve such A I triple E and earlier JEE Main questions on the Bohr model, so let's begin.

0:41Bohr's model: how everything scales

So, first, a quick recap from this chapter. Bohr's model works for hydrogen-like atoms only: one electron going round a nucleus of charge Z times e, where Z is the atomic number and e is the magnitude of the electron's charge. One electron, that is the condition. Okay? So hydrogen has Z equal to one, helium plus has Z equal to two, and lithium two plus has Z equal to three. The small plus signs up top show the charge on each ion, and each of them has just one electron left. Right? Now Bohr's rule. The angular momentum, m times v times r, with m the mass, v the speed and r the radius, equals n times h, over two pi. Here h is Planck's constant, pi is the Greek letter, and h over two pi is written h bar, the h with a little bar across it. And n, the principal quantum number, takes the values one, two, three, and so on. With me? Now put this rule together with the electric pull of the nucleus, as we did earlier in the chapter. The radius r is proportional to n squared over Z, and the speed v is proportional to Z over n. The stretched alpha sign means proportional to, and the small two up top means squared. Clear? Now the period, capital T, is the time for one round. Using the radius and the speed, how does the period depend on n and Z? Very good, one round is two pi r, covered at speed v, so T equals two pi r over v. And n squared over Z, divided by Z over n, gives n cubed over Z squared; the small three means cubed. The frequency, small f, the number of rounds per second, is one over T, so it goes as Z squared over n cubed. So the period goes as n cubed, not as n. Remember this one. Got it?

2:36Energy levels and the ionisation trap

Now the energy. E n, the E with a small n below, is the energy of level n, and it equals minus thirteen point six times Z squared over n squared, in electron volts, the e V on screen. Minus, because the electron is bound to the nucleus, and we must supply energy to pull it free. Okay? So the ionisation energy, the energy needed to remove the electron completely from level n, is the same expression without the minus sign: thirteen point six times Z squared over n squared electron volts. It is the binding energy of the electron in that level. With me? Now an A I triple E question from two thousand two. Thirteen point six electron volts is needed to ionise a hydrogen atom. How much energy is needed to remove the electron from n equal to two? Well, n is doubled, so half the energy, six point eight electron volts, right? Oh no, careful, that is the trap, because the energy goes as one over n squared, not one over n. With Z equal to one, that is thirteen point six over two squared, which is thirteen point six over four, giving three point four electron volts. Half is the trap, and a quarter is the answer. Clear?

3:54Example 1: He+ first excited state ionisation energy (2019)

Now, Example one, from JEE Main twenty nineteen. The question says: a helium plus ion is in its first excited state. Find its ionisation energy. Pause the video and try it first. Done? Okay, let's check. First, the first excited state: is that n equal to one, or n equal to two? It is n equal to two. The ground state is n equal to one, and the first excited state is the first level above it, so n equal to two. And helium plus has Z equal to two. Many students take n equal to one here, so watch out. Okay? Now our formula: thirteen point six times Z squared over n squared. Z squared is four, and n squared is also four. So we get thirteen point six times four, over four. With me? The fours cancel, and the answer is thirteen point six electron volts, the same number as for hydrogen in its ground state! And if you got fifty four point four electron volts, you used n equal to one; as the bracket on screen says, that is the energy needed from the ground state of helium plus. Got it?

5:11Example 2: frequency of revolution in hydrogen n = 2 (2020)

And now Example two, from JEE Main twenty twenty, and yes, this is our opening question. The period of revolution of the electron in the ground state of hydrogen is one point six times ten to the power minus sixteen seconds; the small s on screen means seconds. Find the frequency of revolution in the first excited state. Pause the video and try it first. Done? Okay, let's check. First excited state, so n goes from one to two, as we just saw. Now, our opening answer was that n doubles, so the period doubles. Is that right? No, that is exactly the trap, because the period goes as n cubed over Z squared. For hydrogen, Z is one. So T is proportional to n cubed. Okay? So T two, the T with a small two below, is the period in the first excited state. It is two cubed, which is eight, times T one, the period in the ground state. That is eight times one point six times ten to the power minus sixteen seconds. Right? Eight times one point six gives twelve point eight. So T two is twelve point eight times ten to the power minus sixteen seconds, which is one point two eight times ten to the power minus fifteen seconds. Eight times longer, not twice. With me? Now, many students stop here and write this down. Read the question again. Is this our answer? Not yet, because the question asks for the frequency, not the period. The frequency f two is one over T two, so one over one point two eight times ten to the power minus fifteen. The unit, s to the power minus one, means per second. Okay? One over one point two eight gives zero point seven eight one. So f two is zero point seven eight one times ten to the power fifteen, which is about seven point eight times ten to the power fourteen per second; the wavy equals sign means approximately. So the period became eight times longer, and the frequency became one eighth. That answers our opening question.

7:29Quick revision: r, v, T, f scaling and ionisation energy

Clear? So, quick revision. For hydrogen-like atoms: radius goes as n squared over Z, speed as Z over n, period as n cubed over Z squared, and frequency as Z squared over n cubed. The ionisation energy from level n is thirteen point six times Z squared over n squared electron volts. And the first excited state means n equal to two, and always check whether the question asks for the period or the frequency. Now go and try the practice problems on this page. All the best!