Quantum and Nuclear Physics (B)

Mr. Klapholz Shaker Heights High School Quantum and Nuclear Physics (B) Problem Solving

Problem 1 Review problem. An electron is accelerated through a potential difference of 30 V. • What is its final speed? • How does the energy of the electron compare to the energy of a proton that was accelerated through 30 V? • How does the speed of the electron compare to the speed of the proton?

Solution 1 eV = ½mv2 (1.6 x 10-19)×(30) = ½(9.1x10-31 kg)v2 v = ? v = 3.2 x 106 m s-1 b) They have the same energies because they have the same charge and went through the same voltage difference. c) For a proton to have the same kinetic energy as an electron, the proton must be moving slower.

Problem 2 For a particular metal, high wavelengths do not produce any photoelectrons, and only when the wavelength is dialed down to 300 nm do any electrons appear. [No stopping voltage is applied.] What is the work function for this metal, and why do small wavelengths produce photoelectrons, and large wavelengths do not?

Solution 2 (1 of 2) Small wavelengths are high frequency waves. High frequency is high energy. To produce photoelectrons, the energy of the photons must exceed the work function, so only high-energy photons produce electrons.

Solution 2 (2 of 2) c = fl At the threshold: c = f0l0 f0 = c / l0 = 3.00 x 108 m s-1 ÷ 300 x 10-9 m f0 = 1.00 x 1015 Hz • = hf0 • = (6.6 x 10-34 J s) × (1.00 x 1015 Hz) = 6.6 x 10-34 units? = 6.6 x 10-34 J

Problem 3 What is the de Broglie wavelength of an electron that is moving at 3.2 x 106m s-1?

Solution 3 p = h ÷ l l= h ÷ p l= h ÷ [ mv ] • = (6.6 x 10-34 J s) / [(9.1x10-31 kg) x (3.2x106ms-1)] • = 2.3 x 10-10m

Problem 4 Find the wavelength of light when an electron in hydrogen undergoes a transition from n=2 to n=1.

Solution 4 DE = -3.4 eV – 13.6 eV DE = 10.2 eV We’re going to need this energy in Joules: 10.2 eV ( 1.601 x 10-19 J / 1 eV ) = 1.6 x 10-18 J E = hf f = E / h f = (1.6 x 10-18 J) / (6.6 x 10-34 J s) = 2.4 x 1015 Hz l = c / f l = (3.00 x 108m s-1) ÷ (2.4 x 1015 Hz) l = 1.2 x 10-7 m

Problem 5 In the ground state, how much energy (in eV) does an electron have if it is in a box that is the size of a hydrogen atom 1.0 x 10-10m?

Solution 5 Ek = n2h2 / 8meL2 Ek = 12(6.6x10-34Js)2/[ 8(9.1x10-31kg)12 ] Ek = ? Ek = 6.0 x 10-28 J 6.0 x 10-28 J (1 eV / 1.601 x 10-19 J ) = ? = 3.7 x 10-9eV

Problem 6 The speed of an electron is measured to be 1.00 ± 0.01 x 106m s-1. What is the smallest value possible for the uncertainty in the position of the electron at that time?

Solution 6 (part 1 of 4) The uncertainty in the position comes from the HUP: Dx × Dp ≥ h ÷ 4p. The smallest possible value for the uncertainty in the position is: Dx = h ÷ (4p Dp) So we need the uncertainty in the momentum…

Problem 6 (2 of 4) The ‘quick and dirty’ method for uncertainty goes like this: p = mv = (9.1x10-31kg) x (1.00 x 106m s-1) = 9.1 x 10-25 kg m s-1 The greatest momentum possible: p+ = mv = (9.1x10-31kg) x (1.01 x 106m s-1) = 9.19 x 10-25 kg m s-1 The least possible momentum : p- = mv = (9.1x10-31kg) x (0.99 x 106m s-1) = 9.009 x 10-25 kg m s-1

Solution 6 (3 of 4) p+ - p = 9.19 x 10-25 - 9.1 x 10-25 = ? = 0.09 x 10-25 p - p- = 9.1 x 10-25 – 9.009 x 10-25 = ? = 0.09 x 10-25 So the uncertainty in the momentum is: Dp = 0.09 x 10-25 kg m s-1 And the uncertainty in position is: Dx = h ÷ (4p Dp) = (6.6x10-34Js) ÷ (4p 0.09 x 10-25 kg m s-1)

Solution 6 (4 of 4) And the uncertainty in position is: Dx = ? Dx = h ÷ (4p Dp) = (6.6x10-34Js) ÷ [ (4p)×(0.09x10-25kgms-1) ] = 5.8 x 10-9m

Tonight’s HW: Go through the Quantum and Nuclear section in your textbook and scrutinize the “Example Questions” and solutions. Bring in your questions to tomorrow’s class.

Quantum and Nuclear Physics (B)