(c) An electron and a photon both have an energy of 50 keV....

(c) An electron and a photon both have an energy of 50 keV. Calculate which has the longer de Broglie wavelength? You can assume that the electron energy is non-relativistic. (d) A particular nucleus has an excited state lifetime of 0.250 ns and emits a ?–ray of energy 2.30 MeV when it decays. What is the fractional uncertainty in frequency, ?f / f0 of the ?– rays emitted from these nuclei? (e) Consider a particle bound by a harmonic oscillator potential U(x) = 1/2 Kx^2 (where K is a constant). The wavefunction of the particle is given by: ?(x) = Ae^-bx^2 where A and b are constants and b > 0 (i) Find the normalisation condition for this wavefunction? (ii) Find an expression for the expectation value of the energy. (f) If a normalised wavefunction ? , is an eigenfunction of the momentum operator [p] the momentum p is sharp. Hence: (i) Show that the expectation value of the momentum is the eigenvalue of p (as you would expect) and (ii) show that the uncertainty in the momentum is zero.

Transcript

00:01 All right, so we're going to be dealing with some various quantum mechanical problems.

00:06 The first one is if we have an electron with 50 kiloelectron volts of energy, and we have a photon that is 50 kv.

00:20 Question is, which has a longer debrosialy wavelength.

00:41 And so the debrosia length for light beams is just going to be given by the standard e equals hc over lambda which means that lambda is going to equal hc over e and we can use the fact that h times c in these units is 1 ,242 nanometers.

01:19 And so we can say lambda.

01:22 So this is lambda for the photon is going to equal photon.

01:33 Lambda for the photon is going to equal 1242 over and let's use just kev.

01:46 So 1242 divided by 50 equals 24 .84 nanometers.

01:56 That's because this is in nanometers.

02:04 And so then for the electron, we need to use lambda v equals h over p.

02:25 But we know that the energy, the free energy of our electron, because it's non -relativistic, can be given by p squared over 2m.

02:36 So solving for p, we get p equals square root of 2m times e.

02:41 And using that the mass of an electron is 0 .511 m .e .v.

02:48 Per c squared, we can get that lambda b equals h over square root of two times.

03:04 And so m -e -v to k -e -v, m -e -v is 10 to the 6, k -e -v is 10 to the 3rd.

03:15 So we need to multiply this by 3 or by a thousand to get it in terms of k -e -v.

03:28 K -e -v per c -squared times 50 k -a -v.

03:38 And we can use this c -squared to flip.

03:43 So we can take this c -squared and then pulling it out of our integral or out of our square root.

03:51 It's just a factor of 1 over c, which is on the bottom.

03:55 So then it's put up here.

03:59 Then we get lambda b equals hc over 2 times 511 times 50, which equals once again 1242 over the square root of 2 times 511.

04:23 50, and we can already see this is going to be much smaller.

04:28 But just for completeness, we can work this out.

04:39 And we get 5 .49 nanometers and so comparing the two we can see that the photon has the longer de brochure wave length so next question is if an atom with a half -life of 2 .0 .0 .25 nanosecond emits a photon of 2 .3 m .e .v.

05:54 And our question is, what is the fractional uncertainty in the frequency of this emitted photon? so what we can use is the energy time uncertainty principle, which says that delta e, delta t must always be greater than or equal to h bar over 2.

06:20 So if we say that we want the most, the minimum uncertainty possible, then this becomes inequality.

06:31 And our delta t is going to be 0 .25 nanoseconds.

06:35 So we're going to get delta e times 0 .25 times 10 to the negative 9 seconds equals and then h bar in units of ev is given by 6 .582 e to the negative 16 times 10 of the negative 16 over 2.

07:04 And so just running through these calculations, we can get that the uncertainty in our energy is going to be 1 .37 send to the negative 6 electron volts.

07:23 And so if we use, so for a photon, and we actually used this before in a slightly different form, e equals h new.

07:37 So if we then take our 1 .37 times 10 to the negative 6 electron volts equals 6 .582 times 10 to the negative 16 multiplied by our frequency.

08:00 So that's going to give us a frequency now of new equals 1 .37, 10 to the negative 6 electron volts over 6 .582 times 10 to negative 16 electron volts.

08:27 And so this is just the delta f that we're finding right now.

08:33 So let's actually, i'll call it that instead of calling it new.

08:40 That might get confusing.

08:42 So this is the delta f because this value is delta e.

08:50 So this new here is now going to be delta f in this case.

08:59 So taking these numbers, we can get the delta f 2 .08 times 10 to the.

09:17 9th hertz and so then finding our f not so for f not we can once again use the energy of our photon so 2 .3 mav so 2 .3 that's 10 to the 9th electron volts equals 6 .582 times 10 to the negative 16 so h bar times our f -not 2 .3 e to the 9th divided by 6 .582 e to the negative 16th.

10:19 It's going to give us a very large number of 3 .49 times 10 to the 24th equals f -not.

10:36 So delta f over f not equals in this case taking our delta f two point zero eight times ten to the nine dividing that by three point four nine times ten to the fourth you get five point nine six times ten to the negative sixteen so a very small number...

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