Postulates, Radius and Speed

Transcript

0:00Introduction

Hello everyone! By classical physics, the electron in Rutherford's atom should radiate, spiral inwards, and the atom should collapse. Yet real atoms are perfectly stable. So what rescued the atom? In nineteen thirteen, Niels Bohr made a bold move: only certain orbits are allowed, and in them the electron does not radiate. Today, two equations will give us the radius of every orbit and the electron's speed in it. This comes up often in JEE, so let's begin.

0:31Bohr's postulates (1913)

So, here is the first postulate: an electron can revolve in certain stationary orbits without radiating energy. Stationary does not mean at rest, okay? It keeps moving but does not radiate, so the atom does not collapse. Clear? But which orbits? The second postulate says the angular momentum, capital L, equal to m v r, mass times speed times radius, must be n times h over two pi. Here h is Planck's constant, and n, the principal quantum number, is one, two, three, and so on. The h with a bar, h bar, is short for h over two pi. With me? The third postulate explains the light. When the electron jumps from energy E i to a lower energy E f, shown by the arrow, it emits a photon. The subscripts i and f mean initial and final. Its frequency is nu, the Greek letter like a v, and h nu equals E i minus E f. Got it?

1:34Hydrogen-like atom: one electron, nuclear charge Ze

Hydrogen-like means just one electron, round a nucleus of charge Z e, with Z, the atomic number, and e, the charge of a proton. These results hold only for such atoms, okay? Now, the electric attraction gives the centripetal force: m v squared over r equals Z e squared over four pi epsilon naught r squared. Epsilon naught, the Greek epsilon with a small zero, is the permittivity of free space. Right? Second, Bohr's condition: m v r equals n times h over two pi. Two equations, two unknowns, the speed v and the radius r. With me? Multiply the force equation by r, and divide by the second one. Almost everything cancels, and v n, the speed in the nth orbit, with n as a small subscript, equals e squared over two epsilon naught h, times Z over n. Got it? Put this speed back, and solve for the radius: r n equals epsilon naught h squared over pi m e squared, times n squared over Z, okay? So remember the key parts: Z over n for the speed, and n squared over Z for the radius. Clear?

2:49The numbers to remember

Now put in the constants. The radius is zero point five two nine times n squared over Z, in angstrom, the capital letter with a small circle on top, which is ten to the power minus ten metre. Okay? For hydrogen's first orbit, n and Z are both one, giving zero point five two nine times ten to the power minus ten metre. This is the Bohr radius, written a naught, the letter a with a small zero below it. Right? And the speed is two point one nine times ten to the power six, times Z over n, metres per second, the m s to the power minus one. With me? Now a question for you. Hydrogen's first orbit gives two point one nine times ten to the power six metres per second, and light, c, goes at three times ten to the power eight metres per second. What fraction of c is that? Divide them, and you get about one over one hundred and thirty seven. So v one equals c over one hundred and thirty seven. This number is worth remembering, got it?

3:57How everything scales with n and Z

Now the proportionalities, where the traps live. The symbol like a stretched alpha means proportional to: the radius goes as n squared over Z, and the speed as Z over n, okay? So in a higher orbit, is the electron faster or slower? Yes, it is slower, not faster, because a bigger n means a smaller speed. And a bigger Z gives a smaller orbit, since the nucleus pulls harder. Right? The angular speed, omega, the Greek letter like a curly w, is v over r. Z over n, divided by n squared over Z, gives Z squared over n cubed. The frequency f, revolutions per second, goes the same way. Clear? The period, capital T, is one over the frequency, so it goes as n cubed over Z squared. Watch the powers: three on n, and two on Z, okay? Mixing them up is a common mistake. With me? For hydrogen's ground state, the period is one point five two times ten to the power minus sixteen seconds, and the frequency is six point five eight times ten to the power fifteen hertz, the H z on screen. Right?

5:11The de Broglie picture

Now the de Broglie picture. Two pi r n, the circumference of the nth orbit, equals n times lambda, where lambda, the Greek letter, is the electron's de Broglie wavelength. So how many wavelengths fit round the third orbit? Exactly three wavelengths, because n is three. In the nth orbit, a whole number n of wavelengths fits round the circle.

5:36Example 1: radius of the second orbit of He+

Okay? Now, example one, an easy one. Find the radius of the second orbit of helium plus, the helium ion with a small plus sign up high. Pause the video and try it first. Done? Okay, let's check. First we need the value of Z, okay? Helium plus has one electron, so some students put Z equal to one. Is that right? No, no, careful here, because Z is the nuclear charge, not the number of electrons. Helium plus has one electron, so it is hydrogen-like, and helium's atomic number is two, so Z is two. The second orbit means n is two. With me? Now the radius is zero point five two nine times n squared over Z, in angstrom. That is zero point five two nine times two squared over two, which is four over two. Right? So we get one point zero five eight angstrom, twice the Bohr radius. The wavy equals sign means approximately, so in metres that is about one point zero six times ten to the power minus ten metre. Clear?

6:46Example 2: Be3+ and He+ orbits matching hydrogen n = 1

Now a harder one. Which orbit of beryllium three plus has the same radius as the first orbit of hydrogen? And which orbit of helium plus gives the same speed as the first orbit of hydrogen? Pause the video and try it first. Done? Okay, let's check. For hydrogen's first orbit, n and Z are both one, so n squared over Z and Z over n both equal one. These are our targets. Okay? Now beryllium three plus, with three plus written up high. Its charge is three plus, so is Z equal to three? No, that is the same trap again, because Z is the atomic number, and beryllium's is four; three plus only means three electrons are gone. So n squared over four equals one, and n equals two, not four. The double arrow means implies. Got it? And helium plus, with Z equal to two. For the speed, two over n equals one, so n equals two. Be careful, because n is not squared here. So it is the second orbit in both cases. Right?

7:57Quick revision: mvr = nh/2pi, r and v formulas, the Z trap

So, quick revision. Bohr's condition is m v r equals n times h over two pi, and h nu equals E i minus E f. The radius is zero point five two nine times n squared over Z angstrom, and the speed is two point one nine times ten to the power six, times Z over n, metres per second. These hold for hydrogen-like atoms only, where Z is the atomic number. Now try the practice problems on this page. All the best!