Solving Kinetic Energy Level Problem with Bohr's Rules for B=10T

silverek
Messages
2
Reaction score
0
I am having trouble with this problem,

According to Bohr's quantization rules, the angular momentum of the electron in a given orbit is quantized according to P=nh. Find an equation for the possible kinetic energy levels for the electron orbiting in circles in a magnetic field, B. Calculate the kinetic energy level spacing if B=10T.
 
Physics news on Phys.org
Try it like this: the magnetic field is what generates the centripetal force on the electron, causing it to go in a circle. So:

\frac{mv^2}{r} = qvB

(assuming the electron is traveling perpendicular to the field, which is the case if we want the electron going in a circle)

You can fiddle with this to get an expression for v which you can sub into the kinetic energy equation. You can then introduce the quantisation condition by going back to the equation above, fiddling to get an expression for L = mvr and then replacing it with n\hbar (note that it's h-bar, not just h), then substituting in your energy equation. You will get an equation for kinetic energy that depends on n and is independent of v and r. Indeed, you should get a very familiar expression!

Cheerio!

Kane
 


To solve this problem, we can use the equation for the kinetic energy of a charged particle in a magnetic field, which is given by KE= 1/2mv^2 = 1/2(mv)^2. We know that the angular momentum of the electron in a given orbit is quantized according to P=nh, where n is the principal quantum number and h is Planck's constant.

Substituting this into the equation for kinetic energy, we get KE= 1/2(mv)^2 = 1/2(h/2πr)^2 = (h^2/8π^2mr^2).

Since we are given that B=10T, we can use the equation for the magnetic field strength in terms of the radius of the orbit, r, and the electron's charge and mass, B= mv/2πr. Solving for v, we get v= 2πrB/m.

Substituting this into the equation for kinetic energy, we get KE= (h^2/8π^2mr^2) = (h^2/8π^2m(2πrB/m)^2) = (h^2/8π^2m^2(4π^2r^2B^2/m^2)) = (h^2/32π^2m^2r^2B^2).

Since we are looking for the possible kinetic energy levels, we can use the equation for the energy of an electron in a given orbit, E=-Rhc/n^2, where R is the Rydberg constant and n is the principal quantum number.

Substituting this into the equation for kinetic energy, we get KE= (h^2/32π^2m^2r^2B^2) = -Rhc/n^2.

Solving for n, we get n= √(Rhc/KE).

Now, to calculate the kinetic energy level spacing, we can use the equation ΔKE= KE(n+1)-KE(n), where n is the principal quantum number of the higher energy level and n+1 is the principal quantum number of the lower energy level.

Substituting our previous equation for n into this equation, we get ΔKE= KE(√(Rhc/KE)+1)-KE(√(Rhc/
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

Similar threads

Back
Top