Complete Question:
When specially prepared Hydrogen atoms with their electrons in the 6f state are placed into a strong uniform magnetic field, the degenerate energy levels split into several levels. This is the so called normal Zeeman effect.
Ignoring the electron spin what is the largest possible energy difference, if the magnetic field is 2.02 Tesla?
Answer:
ΔE = 1.224 * 10⁻²² J
Explanation:
In the 6f state, the orbital quantum number, L = 3
The magnetic quantum number, ![m_{L} = -3, -2, -1, 0, 1, 2, 3](https://tex.z-dn.net/?f=m_%7BL%7D%20%3D%20-3%2C%20-2%2C%20-1%2C%200%2C%201%2C%202%2C%203)
The change in energy due to Zeeman effect is given by:
![\triangle E = m_{L} \mu_{B} B](https://tex.z-dn.net/?f=%5Ctriangle%20E%20%3D%20m_%7BL%7D%20%5Cmu_%7BB%7D%20B)
Magnetic field B = 2.02 T
Bohr magnetron, ![\mu_{B} = 9.274 * 10^{-24} J/T](https://tex.z-dn.net/?f=%5Cmu_%7BB%7D%20%3D%209.274%20%2A%2010%5E%7B-24%7D%20J%2FT)
![\triangle E = 6 * 9.274 * 10^{-24} * 2.2\\](https://tex.z-dn.net/?f=%5Ctriangle%20E%20%3D%206%20%2A%209.274%20%2A%2010%5E%7B-24%7D%20%2A%202.2%5C%5C)
ΔE = 1.224 * 10⁻²² J
As a non-metal, since it's a gas
A rock at the edge of a cliff has potential energy because it isn't in motion, but is located in a place that motion could occur in the near future.
Answer:
the potential energy is zero, and the kinetic energy must be maximum
F = 0
Explanation:
In this exercise you are asked to complete the sentences of a simple harmonic movement of a mass-spring system.
In this system mechanical energy is conserved
at the most extreme point the carousel potential energy is
K_e = ½ k x²
the kinetic energy is zero for that stopped.
At the equilibrium point
the spring elongation is x = 0 so the potential energy is zero
and the kinetic energy must be maximum since total energy of the system is conserved
the spring force is
F =- k x
as in the equilibrium position x = 0 this implies that the force is also zero
F = 0
Answer:
Explanation:
Check the attachment for the ray diagram
Ray diagrams are important in locating the position of image of an object. It enables us to determine the nature of image formed by an object in from of a mirror or lens.
When drawing ray diagrams to locate the image I of an object O, there are two important steps to follows:
1. Ray emitted from the tip of an object O incidented on the mirror and parallel to the principal axis OP pass towards the focus F of the mirror after reflection.
2. The ray emitting from the tip of the mirror and passing through the centre of curvature of the mirror C undeflected reflects back on the same line of incidence. (Principle of reversibility).