"Werner Heisenberg was one of the best of those physicists who developed early quantum mechanics. Not only did his work enable a description of nature on the very small scale, it also changed our view of the availability of knowledge. Although he is universally recognized for his brilliance and the importance of his work (he received the Nobel Prize in 1932, for example), Heisenberg remained in Germany during World War II and headed the German effort to build a nuclear bomb, permanently alienating himself from most of the scientific community". (Wikimedia Commons)
We find some problems to essay and prepare your next exam (https://pressbooks.bccampus.ca/collegephysics/chapter/probability-the-heisenberg-uncertainty-principle/).
1: (a) If the position of an electron in a membrane is measured to an accuracy of 1,00 µm, what is the electron’s minimum uncertainty in velocity? (b) If the electron has this velocity, what is its kinetic energy in eV? (c) What are the implications of this energy, comparing it to typical molecular binding energies?
2: (a) If the position of a chlorine ion in a membrane is measured to an accuracy of 1,00 µm, what is its minimum uncertainty in velocity, given its mass is 5,86.10-26 kg? (b) If the ion has this velocity, what is its kinetic energy in eV, and how does this compare with typical molecular binding energies?
3: Suppose the velocity of an electron in an atom is known to an accuracy of 2,0.103 m/s (reasonably accurate compared with orbital velocities). What is the electron’s minimum uncertainty in position, and how does this compare with the approximate 0,1nm size of the atom?
4: The velocity of a proton in an accelerator is known to an accuracy of 0,250% of the speed of light. (This could be small compared with its velocity.) What is the smallest possible uncertainty in its position?
1.a)ΔV= 5,80×10^6 m/s
ResponderEliminarb) Ec=1,53×10^-17 J,es decir, Ec=9,56×10^1eV
2.a)ΔV=9,01×10^1 m/s
b) Ec=2,64×10^-24 J, es decir, Ec=1,65×10^-5 eV
3.Δr =2,9×10^-8 m. Por lo tanto, es 290 veces mayor.
4.Δr =4,22×10^-14 m.
Irene, en el problema 1, apartado a) obtengo cinco ordenes de magnitud menos en la indeterminación de la velocidad del electrón. Por tanto también obtengo distintos valores para los otros dos apartados (había que buscar un valor medio de la energía de enlace para poder establecer comparaciones). En el problema 2, apartado a) me vuelve a pasar los mismo: cinco órdenes de magnitud menos.
EliminarPara los problemas 3 y 4 obtengo el mismo resultado que tú.
Gracias por participar
1:
ResponderEliminara) ΔV=5’80x10^1 m/s
b) Ec= 1’53x10^-27 J; Ec= 2’45x10^-8 eV
2:
a) ΔV=9’01x10^-4 m/s
b) Ec= 2’38x10^-32 J; Ec= 3’81x10^-13 eV
3:
Δr= 2’9x10^-8 m Es 290 veces mayor
4:
Δr= 4’22x10-14 m