Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Here is a quick puzzle. Light falls on a photocell, and we double its frequency while halving its intensity. So the stopping potential doubles, and the photons per second are halved, right? Hold that thought; it is problem ten of today's set. This is Part A of the dual nature problem bank, with eighteen problems at JEE Main level. So, let us sharpen three tools, and their traps.

0:30Tool 1: the photon, in electron volt

Tool number one is the photon. Its energy, capital E, equals h nu, which equals h c over lambda. Here h is the Planck constant, nu, the Greek letter like a v, is the frequency, c is the speed of light, and lambda is the wavelength. Okay? Joules are awkward, so we use the electron volt, the small e and capital V on screen. h times c is twelve hundred and forty electron volt nanometres, n m being the nanometre, and one electron volt is one point six times ten to the power minus nineteen joule, the capital J; the cross means times. Right? So the photon energy in electron volt is twelve hundred and forty over the wavelength in nanometres; the brackets give the units. But if the wavelength is in angstrom, do you still divide twelve hundred and forty by it? No, that is a common slip. Then h c is twelve thousand four hundred electron volt angstrom; the capital A with a small ring on top is the angstrom. Always check the unit first. Clear?

1:32Tool 2: Einstein's photoelectric equation

Tool number two is Einstein's equation. In the figure, a photon of energy h nu, the wavy orange line, gives all its energy to one electron, the dark dot. To leave the metal, it must spend at least phi naught, the work function; phi is the Greek letter drawn as a circle with a line through it, and the small zero is read as naught. The most loosely bound electrons spend just phi naught, so K max, the maximum kinetic energy, equals h nu minus phi naught, which equals e V naught, the electron's charge times the stopping potential. With me? Now tell me, if h nu is less than phi naught, that is the less-than sign you see, but the light is very intense, will any electron come out? No electron at all! One photon gives all its energy to one electron, and if that energy is less than phi naught, the electron cannot leave. More intense light only means more photons, each one still too weak. So below the threshold there is no emission, however intense the light. Right? At the threshold, h nu just equals phi naught. So the threshold frequency, nu naught, is phi naught over h, and the threshold wavelength, lambda naught, is twelve hundred and forty over phi naught in electron volt, in nanometres. Okay? And here is the topper's shortcut: K max in electron volt is the stopping potential in volt, the same number. Got it?

2:55Problem 10: double the frequency, halve the intensity

Back to our puzzle, problem ten. Frequency doubled means each photon carries twice the energy: h nu becomes two h nu, and the arrow means becomes. Intensity halved means the power, capital P, becomes P over two. Okay? So, how many photons arrive every second now? Half as many? No, no, wait, it is one fourth! Photons per second are the power over the energy of one photon, so we get one half over two, which is one fourth. Got it? Now the stopping potential. The photon energy doubled, so it simply doubles too, right? Oh no, careful, that is the trap: the work function is subtracted first. The new K max is two h nu minus phi naught, which is two times, h nu minus phi naught, plus phi naught. With me? But h nu minus phi naught is the old K max. So the new K max is twice the old one, plus phi naught. Puzzle solved: the stopping potential more than doubles, and the photons per second fall to one fourth. Clear?

4:01Tool 3: the de Broglie wavelength

Tool number three is for matter waves. Every moving particle of momentum p has a de Broglie wavelength, lambda equals h over p. With kinetic energy K and mass m, it is h over the square root of two m times K; the tick-shaped sign is the square root. For a charge q accelerated from rest through a potential difference V, it is h over the square root of two m q V, at non-relativistic speeds. Okay? For an electron only, it becomes one point two two seven over root V, in nanometres. Now, two particles have the same kinetic energy, and one is heavier. Which has the longer wavelength? The lighter one, yes, exactly! For the same K, lambda is proportional to one over root m; the open loop sign means is proportional to. Right? So how do we know electrons are waves? The figure shows the Davisson Germer experiment: electrons from a gun strike a nickel crystal, and a movable detector measures the intensity at each scattering angle, theta. At fifty four volts, a peak appears at fifty degrees, the small circle meaning degrees, and the measured wavelength, zero point one six five nanometres, matches one point two two seven over root fifty four, which is zero point one six seven.

5:19Example 1: K max of photoelectrons, 350 nm light on sodium

Got it? Example one is problem five of the set. Light of three hundred and fifty nanometres falls on sodium, with work function phi naught equal to two point seven five electron volt. Find the maximum kinetic energy of the photoelectrons. Pause the video and try it first. Done? Okay, let's check. The photon energy is twelve hundred and forty over three hundred and fifty, about three point five four electron volt; the wavy equals sign means approximately. Now, before we subtract, what must we check? Yes, exactly, the threshold! Three point five four is greater than two point seven five, that is the greater-than sign on screen, so electrons are emitted. Right? So K max equals three point five four minus two point seven five, which is zero point seven nine electron volt.

6:12Example 2: max speed of photoelectrons when V0 = 2.5 V

Okay? Now a harder one. Example two is problem six of the set. The stopping potential in an experiment is two point five volts. Find the maximum speed of the photoelectrons. Pause the video and try it first. Done? Okay, let's check. Since e V naught equals K max, K max is two point five electron volt. Can we put two point five straight into half m times v squared? No, no, wait, a speed in metres per second needs the energy in joules. Two point five times one point six times ten to the power minus nineteen gives four point zero times ten to the power minus nineteen joule. Right? Now, half m times v max squared equals K max, where v max is the maximum speed, and m, the electron's mass, is nine point one times ten to the power minus thirty one kilogram, the k g on screen. With me? So v max is the square root of, two times four point zero times ten to the power minus nineteen, over nine point one times ten to the power minus thirty one, which is root of eight point eight times ten to the power eleven. Okay? Take the root, and v max is about nine point four times ten to the power five metres per second, the m s to the power minus one on screen. So stay in electron volt, and convert to joules only when a speed is needed. Clear?

7:34Quick revision: photon energy, Einstein equation, de Broglie

So, quick revision. Photon energy in electron volt is twelve hundred and forty over lambda in nanometres, or twelve thousand four hundred for angstrom. K max equals h nu minus phi naught, which equals e V naught; subtract the work function first. Matter waves: h over root two m q V, and for an electron, one point two two seven over root V, in nanometres. Now attempt all eighteen problems of Part A as a test, then check the solutions. You have the tools now, so go for it. All the best!