Mixed Problems: Set 1 (Particles, Photons and the Photoelectric Effect)

Transcript

0:00Introduction

Hello everyone! So, one small puzzle before we start. I shine light of four hundred nanometres on a metal, and measure the stopping potential. Then I switch to three hundred nanometres. By how much does the stopping potential change? Oh, and I will not tell you the metal. No name, no work function, nothing. Impossible, right? Not at all! By the end of this class, you will answer it in one line. This is our first mixed problem set for JEE Main: particles, photons and the photoelectric effect. And look, these are the kinds of questions JEE Main often asks. So, let's begin.

0:42Particles: read A, Z and the charge

So, first tool: counting particles. Look at this symbol: potassium, K, with thirty nine on top, nineteen at the bottom, and a small plus sign. The top number is A, the mass number, and the bottom number is Z, the atomic number, okay? Now tell me: how many neutrons does it have, and how many electrons? Neutrons first. N, the number of neutrons, equals A minus Z: thirty nine minus nineteen, that is twenty. Easy, isn't it? Now the electrons. Nineteen protons, so nineteen electrons, right? No, no, careful! That is the trap. Look at that plus sign: it means one electron has been removed. So nineteen minus one: eighteen electrons. Twenty neutrons, eighteen electrons. Clear? Next, find the number of protons in four point four grams of carbon dioxide, C O two; the small two means two oxygen atoms. Moles first, always. The molar mass is forty four, so four point four by forty four: zero point one mole of molecules. Alright, now tell me: how many protons sit in one carbon dioxide molecule? Twenty two, yes! Carbon has six protons, and each of the two oxygens has eight. Six plus two times eight: twenty two. So zero point one times twenty two: two point two N A; N with a small A is Avogadro's number. That is about one point three two times ten to the power twenty four protons; the wavy equals sign means approximately. Moles, times the count in one molecule. That is the whole tool.

2:25Photons: the 1240 shortcut

Now, second tool: photons. Remember the topper's shortcut from chapter two: E in electron volts, written e V, equals twelve hundred and forty over lambda in nanometres. Lambda is the wavelength, and the brackets give the units: e V for electron volts, n m for nanometres. And one mole of photons, each of E electron volts, carries ninety six point five times E kilojoules, written k J on screen. Okay? Now, find the energy of one mole of photons of wavelength three hundred nanometres. Twelve hundred and forty over three hundred: four point one three electron volts per photon. So the answer is four point one three, right? Hold on! That is for one photon, and in electron volts. The question wants one whole mole. So four point one three times ninety six point five: about three hundred and ninety nine kilojoules. Next, the number of photons per second from a two watt source of wavelength five hundred nanometres. N equals P lambda over h c: P is the power, h is Planck's constant, and c is the speed of light. So two, times five times ten to the power minus seven, the wavelength in metres, over one point nine nine times ten to the power minus twenty five, which is h times c. Careful with units here: the wavelength must be in metres, not nanometres. Got it? That gives about five point zero times ten to the power eighteen photons every second. Just think: one small two watt lamp, and such a huge number of photons!

4:06Photoelectric effect: pay the work function first

Now the third tool: the photoelectric effect. Einstein's equation: K max, the maximum kinetic energy of the photoelectrons, equals h nu minus phi naught. Nu, the Greek letter like a v, is the frequency, so h nu is one photon's energy. Phi naught, Greek phi with a small zero, is the work function. Or, h times, in brackets, nu minus nu naught, the threshold frequency. And e V naught equals K max, where V naught is the stopping potential and e is the electron's charge. With me? Now a common JEE pattern. Light of frequency two nu naught falls on a metal of threshold frequency nu naught; the fastest electrons have kinetic energy K one, K with a small one. With three nu naught, it is K two. We want the ratio K one to K two; the colon means ratio. Frequencies are two to three, so the energies are also two to three, right? No, no, careful! The metal takes its share first. K one is h times, in brackets, two nu naught minus nu naught: just h nu naught. K two is h times, in brackets, three nu naught minus nu naught: two h nu naught. So K one to K two is one to two, not two to three. Look, the frequency went only from two nu naught to three nu naught, but the kinetic energy doubled! Always subtract the work function before you compare. Remember that!

5:43Example 1: stopping potential at 310 nm, work function 2 eV

Now, example one. Here, a metal of work function two electron volts is lit by light of wavelength three hundred and ten nanometres. Find the stopping potential. Pause the video, and try it first. Done? Okay, let's check. The photon energy, with our shortcut: twelve hundred and forty over three hundred and ten. Exactly four electron volts. So the stopping potential is four volts, right? No, no, wait! The work function is paid first: our third tool. K max equals four minus two: two electron volts. And e V naught equals K max; the double arrow means this gives. So V naught is two volts. Topper's trick: K max in electron volts and V naught in volts are always the same number. Simple as that!

6:36Example 2: change in stopping potential from 400 to 300 nm

Now the harder one, and it is our opening puzzle. By how much does the stopping potential for a metal change, when the wavelength of the light is changed from four hundred nanometres to three hundred nanometres? No metal named, no work function given. Pause and try. Hint: write V naught for both wavelengths. Done? Okay, let's check. Put our shortcut into Einstein's equation, and use example one's trick. With lambda in nanometres and phi naught in electron volts, V naught equals twelve hundred and forty over lambda, minus phi naught, in volts. Now write it twice. At four hundred nanometres, V one, V with a small one, is twelve hundred and forty over four hundred, minus phi naught. At three hundred, V two is twelve hundred and forty over three hundred, minus phi naught. Now subtract them. What happens to phi naught? Yes, exactly! It cancels. Same metal, same phi naught, gone! As long as both wavelengths eject electrons, the metal does not matter. Delta V, with this triangle, the Greek capital delta, means the change in stopping potential. It equals V two minus V one: twelve hundred and forty, times, in brackets, one by three hundred minus one by four hundred. That is the secret of our puzzle: we never needed the metal! One by three hundred minus one by four hundred is one by twelve hundred. So delta V is twelve hundred and forty over twelve hundred: about one point zero three volts. And the direction: shorter wavelength means a more energetic photon, so the stopping potential rises, by about one point zero three volts. Puzzle solved, in one line. Perfect!

8:27Quick revision: N = A - Z, 1240 over lambda, work function

So, quick revision. Particles: N equals A minus Z, and adjust the electrons for the charge. Photons: twelve hundred and forty over lambda in nanometres gives electron volts, and one mole carries ninety six point five times E kilojoules. Photoelectric effect: pay the work function first; and in a change of stopping potential, the work function cancels. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!