Set 2

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Picture a beam of electrons, each carrying thirteen point zero electron volts, hitting hydrogen atoms in the ground state. Thirteen is so close to thirteen point six, the energy that frees the electron completely. So the atoms must climb very high and give lots and lots of spectral lines, right? Hold on to your answer. Today is Set two of the mixed problems on atoms and nuclei: levels, series, and the nucleus. These come up often in JEE Main, so let's begin.

0:35Tool 1: the energy of level n

So, tool one. The figure shows the levels of hydrogen: each horizontal line is a level, labelled by n, the principal quantum number, with its energy in electron volts, e V, on the right. The dashed line at zero is n equal to infinity, the sideways eight. On screen, E n, with the small n below, equals minus thirteen point six, times Z squared over n squared, electron volts, where capital Z is the atomic number. It holds for hydrogen-like atoms only, with one electron. Okay? To free the electron, we must lift it up to zero. So the ionisation energy from level n is thirteen point six Z squared over n squared electron volts, the same number without the minus sign. Right? Now a problem from this set. In a hydrogen-like ion, the level n equal to three has energy minus thirteen point six electron volts, just like the ground state of hydrogen. So E three, minus thirteen point six times Z squared over nine, equals minus thirteen point six. Which ion is it? Yes, exactly, Z squared over nine must be one, so Z squared is nine, and Z is three. That is lithium two plus, L i with a small two plus up top. With me? Its ground state has n equal to one, so E one is minus thirteen point six times nine, the cross meaning times. That gives minus one hundred and twenty two point four electron volts, nine times hydrogen's value.

2:09Tool 2: spectral series

Clear? Now tool two, the lines. Each arrow in the figure is the electron falling from an upper level n two to a lower level n one, the small numbers below being labels, and one photon comes out. Its wavelength, lambda, the Greek letter on screen, obeys the Rydberg formula: one over lambda equals capital R, the Rydberg constant, times Z squared, times the bracket, one over n one squared, minus one over n two squared. And one over R is ninety one point two nanometres, the n m on screen. Okay? Each series is named by the level it lands on, n one. In the figure, the dark blue arrows land on level one, the Lyman series; the orange ones on level two, Balmer; and the black ones on level three, Paschen. Got it? Now from this set: a Lyman line of hydrogen has wavelength one hundred and two point six nanometres. Lyman, so n one is one, and Z is one. So one minus one over n squared equals one over R lambda, that is ninety one point two over one hundred and two point six, which gives zero point eight eight nine. From which level does the electron fall? One over n squared is zero point one one one, that is one ninth. So n squared is nine, and n is three: the jump from three to one. Right? Now the shortest wavelength of a series, called its limit. The bigger the jump, the shorter the wavelength, so which jump gives the limit? Yes, exactly, the jump from n two equal to infinity, so one over n two squared becomes zero. Then lambda limit, with the word limit below, equals n one squared over R Z squared. Clear? So the Brackett series, landing on n one equal to four, has its limit at sixteen over R for hydrogen. Sixteen times ninety one point two gives one thousand four hundred and fifty nine nanometres. With me?

4:10Tool 3: the size of a nucleus

Now one from the nucleus. Here capital R is the radius of the nucleus, not the Rydberg constant. It equals R naught, times A to the power one third, where capital A is the mass number, the number of nucleons. R naught, the R with a small zero below, is one point two femtometres, the f m on screen. Okay? From this set: how many nucleons are there in a nucleus of radius six point zero femtometres? So A to the power one third is six point zero over one point two, which gives five. So the nucleus has five nucleons, right? Be careful here, because five is only the cube root of the mass number. So A is five cubed, which gives one hundred and twenty five nucleons. The mass number goes as the cube of the radius.

5:02Example 1: energy to ionise He+ from n = 3

Got it? Now, Example one, the easier one. Find the energy needed to ionise a helium plus ion, H e with a small plus up top, that is in the state n equal to three. Pause the video and try it first. Done? Okay, let's check. Ionisation from level three is thirteen point six over nine, which gives one point five one electron volts, right? That would be the answer for hydrogen, so be careful here. Helium plus has Z equal to two, so thirteen point six times Z squared is thirteen point six times four, which gives fifty four point four electron volts. Okay? Now divide by n squared, which is nine. Fifty four point four over nine gives six point zero four electron volts. Forget Z squared, and your answer is four times too small. Clear?

5:59Example 2: lines emitted when 13.0 eV electrons hit hydrogen

Now Example two, and this is our opening puzzle. An electron beam of energy thirteen point zero electron volts bombards hydrogen atoms in the ground state. How many lines can be emitted? Pause the video and try it first. Done? Okay, let's check. Remember, an electron, unlike a photon, can give up any part of its energy. So it can lift the atom to any level needing at most thirteen point zero electron volts. Right? Level one to level four, as the arrow shows, needs thirteen point six times the bracket, one minus one over sixteen, which gives twelve point seven five electron volts. That is less than or equal to thirteen point zero, as the sign says, so level four is reached. Okay? Now level five. Thirteen point zero is so close to thirteen point six, so surely level five is easy, right? Let us calculate instead of guessing. Thirteen point six times one minus one over twenty five gives thirteen point zero six electron volts, and the sign says greater than thirteen point zero. So level five is out of reach, missed by just six hundredths of an electron volt. With me? So the atoms reach n equal to four at most. Many atoms, so the number of lines is n times n minus one, over two: four times three over two, which gives six lines. Not lots of lines, just six, and that solves our puzzle. Got it?

7:33Quick revision: levels, Rydberg, nuclear radius, line count

So, quick revision. Level n has energy minus thirteen point six Z squared over n squared electron volts, and ionising from it needs the same, without the minus sign. One over lambda is R Z squared times the bracket, and each series is named by the level it lands on. For a nucleus, R is R naught times A to the power one third, so the mass number is R over R naught, cubed. And an electron beam lifts atoms to every level within its energy, giving n times n minus one, over two, lines. Now go and try the rest of this set on this page. All the best!