Overview

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Take hydrogen atoms in the ground state, and try to excite them. Send in photons, and a photon can excite an atom only if its energy matches the gap between two levels exactly. Now send in electrons, and any electron with more than ten point two electron volts can excite it. So why is the photon so fussy, while the electron is not? Today we see how to use this unit's problem bank, settle this puzzle with one of its concept checks, and solve two of its problems. Questions on the Bohr model come up often in JEE, so let's begin.

0:40How to use this chapter

So, what is in this chapter? Part A has sixteen solved problems, at JEE Main level. They range from Rutherford and Bohr to binding energy and fission. With me? Then Part B has sixteen harder problems, at JEE Advanced level. Okay? And Part B also brings in X-rays and radioactivity. Right? Part C has twelve concept checks, the why questions, like our photon puzzle. Then come fourteen practice problems to solve alone. Got it? Now tell me, when should you read a solution? Straight away, or only after you have tried it yourself? Yes, exactly, only after you have tried it yourself! The course says it clearly: attempt each part as a test, and only then read the solutions. Clear?

1:33Part C: why is the photon so fussy?

Now a taste of Part C, and our puzzle. The figure shows the energy levels of hydrogen: n equals one at the bottom, and the dashed line at the top is zero, for n equals infinity, the sideways eight. The arrows down are the Lyman series, ending on n equals one, the Balmer series, ending on n equals two, and the Paschen series, ending on n equals three. Beside them is the Rydberg formula for one over lambda, the wavelength, with R, the Rydberg constant, and n one, the lower level, below n two. Today we need only the energies. Okay? Here is the energy formula. E n, the E with a small n below, equals minus thirteen point six times Z squared, over n squared, in electron volts, written e V on screen. Here n is the principal quantum number, and capital Z is the atomic number. It holds only for hydrogen-like atoms, with a single electron. Right? For hydrogen, Z is one. So E one, with a small one below, is minus thirteen point six, and E two is minus three point four zero electron volts. You can see them on the figure too. With me? So tell me, what is the least energy that lifts the electron from n equals one to n equals two? Yes, exactly, it is ten point two electron volts! E two minus E one is minus three point four zero plus thirteen point six, which gives ten point two. This is the first excitation energy of hydrogen. Okay? Now think about the photon. It cannot be split; it is absorbed whole, or not at all. So to excite the atom, its energy must fit a gap between two levels exactly. Note, a photon above thirteen point six electron volts can instead ionise the atom, but that is not excitation. Got it? But an electron is different. It can give up just part of its kinetic energy to the atom, and keep the rest. Clear? Back to our puzzle. Which electrons can excite a hydrogen atom in the ground state? That is correct, any electron above ten point two electron volts can do it! It hands over ten point two to the atom and carries away the rest. The photon has no such choice, so our puzzle is settled. Right?

3:52Example 1: first excitation energy of Be3+

Now, Example one, an easy one from Part A, to warm up. Find the first excitation energy of beryllium three plus; the three plus at the top right means three electrons are gone. Pause the video and try it first. Done? Okay, let's check. Beryllium has Z equals four, so a neutral atom has four electrons. Three are gone, so one is left, and one electron means hydrogen-like. Our formula applies. Right? Now, what does first excitation energy mean? From which level, to which level? That is right, it is from the ground state, n equals one, to the first excited state, n equals two. The arrow on screen means goes to. Okay? The energy needed is the gap, delta E; the little triangle is the Greek capital delta, meaning a change. Delta E equals thirteen point six times Z squared, times bracket one over one squared, minus one over two squared, in electron volts. With me? Z squared is sixteen, and the bracket is one minus one quarter, which is three quarters. Thirteen point six times sixteen times three quarters gives one hundred and sixty three point two electron volts. Now, watch out for a trap here. Some students write thirteen point six times sixteen, which is two hundred and seventeen point six, and stop. Is that the first excitation energy? No, it is not, and that is the trap: two hundred and seventeen point six electron volts is the ionisation energy, the energy to remove the electron completely from n equals one. For excitation, you need the three quarters too. Clear?

5:36Example 2: photons emitted by He+ after absorbing 48.36 eV

Now Example two, a harder one from Part B, at JEE Advanced level. Helium plus ions, the plus at the top right meaning one electron is lost, are in the ground state. They absorb photons of forty eight point three six electron volts. Find the energies of the photons they can then emit. Pause the video and try it first. Done? Okay, let's check. Helium plus has Z equals two and one electron, so it is hydrogen-like. Z squared is four, so E n equals minus fifty four point four over n squared. So E one is minus fifty four point four, and E two is minus thirteen point six electron volts. Okay? The photon is absorbed whole, remember Part C? So the ion's energy becomes minus fifty four point four plus forty eight point three six, which is minus six point zero four electron volts. Right? Now, which level has this energy? Let us solve it: fifty four point four over n squared equals six point zero four, so n squared is nine, and n is three. The double arrow on screen means which gives. So the photon fitted the gap from one to three exactly; that is why it was absorbed. With me? Now the ions come back down. Just one photon, of forty eight point three six electron volts, straight back down, right? Not quite, and that is exactly the trap in this question. From n equals three, an ion can jump straight to one, or go from three to two, and then from two to one. Across many ions, all three jumps happen, so we get three lines. Clear? Three to one gives back the same forty eight point three six electron volts. For three to two, take E three minus E two: minus six point zero four, minus bracket minus thirteen point six, which gives seven point five six electron volts. Got it? And for two to one: minus thirteen point six, minus bracket minus fifty four point four, gives forty point eight electron volts. Quick check: seven point five six plus forty point eight is forty eight point three six, the same energy by the two step route. Very good, problem done!

7:54Quick revision: En formula, photon vs electron, line count

So, quick revision. Part A is JEE Main level, Part BJEE Advanced, and Part C checks the ideas; attempt each as a test first. E n equals minus thirteen point six times Z squared over n squared electron volts, for hydrogen-like atoms only, and first excitation means n equals one to two. To excite an atom, the photon is absorbed whole and must fit a gap exactly, while an electron can give up part of its energy. And many atoms in level n give n times n minus one, over two, different lines, so three lines from n equals three. Now open Part A, solve it like a test, and go right up to the practice problems. All the best!