IIT-JEE Questions (1992–2002): Spectra

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. In hydrogen, the jump from level four down to level three gives a Paschen line, in the infrared. But in an IITJEE question from two thousand one, the same jump, four to three, gives ultraviolet. How is that possible? Example one will answer it. Today we solve the IITJEE questions on spectra from nineteen ninety two to two thousand two. Ideas like these come up often, so let's begin.

0:33Levels, jumps and the photon

So, first, our toolkit; the diagram shows the energy levels of hydrogen. E n, with the small n below, is the energy of level n: minus thirteen point six times Z squared over n squared, in electron volts, the e V on screen. Here n is the principal quantum number, and capital Z is the atomic number. This holds only for one-electron atoms, like hydrogen and hydrogen-like ions. Okay? Now, the electron falls from an upper level, n two, to a lower level, n one. Tell me, how much energy does the photon carry away? Yes, exactly, the difference between the two levels. h nu equals E n two minus E n one, with n two and n one written small below. That is thirteen point six times Z squared, times the bracket, one over n one squared, minus one over n two squared. Here h is Planck's constant, and nu, the Greek letter like a v, is the frequency. Clear? Now, hydrogen-like ions. Every energy is multiplied by Z squared, and every wavelength is divided by Z squared; those are the times sign and the division sign. This line will answer our puzzle. Got it? Now, a phrase that trips many students. When a question says the first excited state, which value of n is that? It is n equals two. If you said n equals one, oh no, careful, that is the ground state itself! So the first excited state is n equals two, and the second excited state is n equals three. With me?

2:10Three tools for these questions

Now, three tools these questions use. First, when an atom emits two photons one after the other, their energies add up to the total drop, from the starting level to the final level. Energy is conserved, right? Second, lambda in nanometres equals twelve hundred and forty over delta E in electron volts. Lambda, the Greek letter, is the wavelength, and delta E, with the little triangle, is the energy gap. So less energy means a longer wavelength. Okay? Now tell me, ultraviolet or infrared: which one carries more energy in each photon? Ultraviolet, yes, because it has the short wavelength and the large energy gap. Infrared has the long wavelength and the small gap. Clear? And third, a topper's trick. When two equations in n and Z start from the same level n, subtract them. The one over n squared terms cancel, and Z comes out straight away. Got it?

3:11Example 1: which jump gives infrared if 4 to 3 gives UV

Now, remember our puzzle? Here is example one, from IITJEE two thousand one. The transition n equals four to three, in a hydrogen-like atom, gives ultraviolet radiation; the arrow means the jump from four down to three. Which of the transitions two to one, three to two, four to two, and five to four gives infrared radiation? Pause the video and try it first. Done? Okay, let's check. Infrared carries less energy than ultraviolet, so the infrared jump must give less energy than four to three; the double arrow means implies. Right? But wait, the question never gives us the value of Z at all. Do we need it to compare these jumps? No, because every jump has the same thirteen point six times Z squared in front, so we compare only the brackets. The diagram shows hydrogen; our ion has the same pattern, only scaled up. For four to three, one ninth minus one sixteenth is seven over one hundred and forty four, about zero point zero four nine; the wavy equals sign means approximately. Okay? Next, two to one gives zero point seven five, three to two gives zero point one three nine, and four to two gives zero point one eight eight. All three are bigger, so they carry more energy and are ultraviolet too. Clear? Only five to four is left. Is its bracket smaller than seven over one hundred and forty four? Let's see. One sixteenth minus one twenty fifth is nine over four hundred, which is zero point zero two two five. Yes, it is smaller, less than half. With me? So, of the four options, only five to four has a smaller gap than four to three, and the answer is five to four. And the puzzle? In hydrogen, four to three is an infrared Paschen line, but every energy is multiplied by Z squared, so with a large Z the same jump moves into the ultraviolet. The order of the lines stays the same. Well done!

5:14Example 2: find n and Z from two pairs of photons

Now a harder one, example two, from IITJEE nineteen ninety four. We have a hydrogen-like atom of atomic number Z, in an excited state n. It can reach the first excited state by emitting, one after the other, two photons of ten point two and seventeen point zero electron volts, or the second excited state by emitting two photons of four point two five and five point nine five electron volts. Find n and Z, okay? Pause the video and try it first. Done? Okay, let's check. The first excited state is n equals two, and the second is n equals three. So, how much energy does the atom give out on its way down to n equals two? Yes, exactly, they add up: ten point two plus seventeen point zero is twenty seven point two electron volts. So for the first excited state, n equals two, thirteen point six times Z squared, times one quarter minus one over n squared, equals twenty seven point two. Okay? For the second excited state, n equals three, four point two five plus five point nine five is ten point two. So the bracket becomes one ninth minus one over n squared, and the right side is ten point two. Right? Now we have two equations, with both n and Z unknown. Remember the topper's trick? What should we do here? Yes, exactly, subtract the second equation from the first, and the one over n squared terms cancel. Thirteen point six times Z squared, times one quarter minus one ninth, equals twenty seven point two minus ten point two, which is seventeen. Clear? One quarter minus one ninth is five over thirty six; the dot means times. Seventeen times thirty six is six hundred and twelve, and thirteen point six times five is sixty eight. Six hundred and twelve over sixty eight gives nine, so Z squared is nine, and Z equals three. Got it? Put Z squared equal to nine in the first equation: thirteen point six times nine is one hundred and twenty two point four. Twenty seven point two over that gives two ninths, so one over n squared is one quarter minus two ninths, which is one over thirty six. With me? So n squared is thirty six, and n equals six, with Z equal to three. Quick check with the second equation: one hundred and twenty two point four, times one ninth minus one over thirty six, gives ten point two. Well done!

7:47Quick revision: Z squared, photons add, subtract equations

So, quick revision. h nu equals thirteen point six times Z squared times the bracket, and every line of the same atom has the same Z squared. The first excited state is n equals two, and photons in succession add up. Less energy means a longer wavelength, and with two equations in n and Z, subtract. Now try the other IITJEE questions on this page. All the best!