IIT-JEE Questions (1982–1990): Spectra

Transcript

0:00Introduction

Hello everyone! Here is a strange little puzzle. We shine light on a gas of atoms, and every photon going in carries two point seven electron volts. But some of the photons coming out carry more than that! So where does the extra energy come from? Hold on to that question. Today we solve the old IITJEE questions on spectra, from nineteen eighty two to nineteen ninety, and this puzzle is one of them, so let's begin.

0:28Energy levels of one-electron atoms

First, the diagram shows the energy levels of hydrogen, with zero energy at the dashed line on top, n equals infinity. 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, and this holds only for one-electron atoms. Okay? Now the electron falls from an upper level, n two, to a lower level, n one; the small numbers below the n tell them apart. How much energy does the photon carry away? Yes, exactly, it carries the difference of the two levels. h nu equals thirteen point six times Z squared, times bracket one over n one squared, minus one over n two squared, in electron volts; h is Planck's constant, and nu, the Greek letter like a v, is the frequency. The diagram writes the same jump as one over lambda equals capital R times Z squared times that bracket, where lambda is the wavelength and capital R is the Rydberg constant. The lower level, n one, names the series on the diagram: one for Lyman, two for Balmer, three for Paschen. Clear? Now a handy shortcut. Lambda in nanometres equals one thousand two hundred and forty over delta E in electron volts; capital delta, the little triangle, means change, so delta E is the energy of the jump. Got it? And one number for today. Lithium two plus, written with a two plus up top, is doubly ionised lithium, with just one electron left. Its atomic number is three, so Z squared is nine, and thirteen point six times nine gives one hundred and twenty two point four electron volts. With me?

2:14Absorbing light, then counting the lines

Next, absorption. Light is absorbed only if the photon energy, h nu, fits the gap between two levels exactly, or if it exceeds the ionisation energy and knocks the electron out. Okay? The excited atoms then fall back, and every jump gives a line. So, if many atoms reach level n, how many different lines can we see? Every pair of levels gives one line, and from n levels we can pick n times bracket n minus one, over two, pairs. Right? But careful, one single atom is different: it gives the most photons when it drops one level at a time, so it gives at most n minus one photons. Spectra come from many atoms, so there we use the first formula. Clear? Now hydrogen-like ions, which also have one electron. Every energy is multiplied by Z squared, and every wavelength is divided by Z squared; see the times sign and the division sign. Okay?

3:12Example 1: Li2+ from orbit 1 to 3, wavelength and lines

Alright, example one, from IITJEE nineteen eighty five. In part A, find the wavelength needed to excite the electron of lithium two plus from the first to the third Bohr orbit. And in part B, how many lines are seen in the emission spectrum of the excited system? Pause the video and try it first. Done? Okay, let's check. For orbit one to orbit three, many students write thirteen point six times eight over nine, which is twelve point zero nine electron volts, as in the badge. Is that right? No, careful, that is the trap, because twelve point zero nine electron volts is the gap for hydrogen. For lithium two plus, Z squared is nine, so we start from one hundred and twenty two point four. Delta E is one hundred and twenty two point four times bracket one minus one ninth, that is, times eight over nine, which gives one hundred and eight point eight electron volts. Got it? Now the wavelength. One thousand two hundred and forty over one hundred and eight point eight, in nanometres, gives about eleven point four nanometres; the wavy equals sign means approximately. Right? Now part B, the lines. One ion in orbit three gives at most two photons, three to two, then two to one. So the answer is two lines, isn't it? Not quite, because the system has many lithium ions, not just one. Some fall from three to one directly, and others go three to two, then two to one. So all three pairs show up, one for each arrow on screen, and three times two over two gives three lines. Very good, that settles part B.

4:55Example 2: six lines after 2.7 eV, find level B and E0

Now our opening puzzle, example two, from IITJEE nineteen eighty nine. We have a gas of identical hydrogen-like atoms, some in the ground level, called A, some in an excited level, called B, and none in any other level. They absorb light of photon energy two point seven electron volts, and then emit photons of only six different energies: some equal to two point seven electron volts, some more, and some less. Find the quantum number of B, the ionisation energy, and the largest and smallest photon energies. Pause the video and try it first. Done? Okay, let's check. Six different photon energies means six lines. So tell me, up to which level did the atoms go? Yes, exactly, the atoms went up to level four, because n times n minus one over two equals six gives n equals four; the double arrow means implies. Clear? So the two point seven electron volt photons lift atoms from level B up to level four. Now, is B level three, just one step below the top? No, B cannot be level three. If B were level three, the two point seven electron volt line would be the jump from four to three, the smallest gap of all, and nothing could come out with less energy. But some photons do carry less, so B is level two, and the jump is two to four. With me? We do not know Z, so let us write the levels as E n equals minus E naught over n squared; E naught, the E with a small zero below, is the energy needed to free the electron from the ground level. So E naught is the ionisation energy. Okay? For the jump two to four, E naught times bracket one quarter minus one sixteenth, which is three E naught over sixteen, equals two point seven. So E naught is fourteen point four electron volts, and that is the ionisation energy. Right? Now the biggest photon comes from the biggest jump, four to one. Fourteen point four times bracket one minus one sixteenth, that is, times fifteen over sixteen, gives thirteen point five electron volts. And our puzzle is solved: no energy was created, because the atoms in level B were already excited, and falling to the ground level releases that stored energy. Clear? And the smallest photon comes from the smallest gap, four to three. Fourteen point four times bracket one ninth minus one sixteenth, that is, times seven over one hundred and forty four, gives zero point seven electron volts. Well done, the whole question is solved.

7:31Quick revision: Z squared scaling and n(n-1)/2 lines

So, quick revision. For hydrogen-like ions, energies are multiplied by Z squared, so for lithium two plus, start from one hundred and twenty two point four electron volts. Many atoms at level n give n times n minus one over two lines, but one atom gives at most n minus one photons. And in puzzles, count the lines to find the top level, then use the clues for the lower one. Now try the other old IITJEE questions on this page. All the best for your preparation.