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, 
The change in energy due to Zeeman effect is given by:
Magnetic field B = 2.02 T
Bohr magnetron, 
ΔE = 1.224 * 10⁻²² J
A 100 g cart is moving at 0.5 m/s that collides elastically from a stationary 180 g cart. Final velocity is calculated to be 0.25m/s.
Collision in which there is no net loss in kinetic energy in the system as a result of the collision is known as elastic collision . Momentum and kinetic energy both are conserved quantities in elastic collisions.
Collision in which part of the kinetic energy is changed to some other form of energy is inelastic collision.
For an elastic collision, we use the formula,
m₁V₁i+ m₂V₂i = m₁V1f + m₂V₂f
For a perfectly elastic collision, the final velocity of the 100g cart will each be 1/2 the velocity of the initial velocity of the moving cart.
Final velocity = 0.5/2
=0.25 m/s.
To know more about elastic collision, refer
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Answer:
W = 3/2 n (T₁- T₂)
Explanation:
Let's use the first law of thermodynamics
ΔE = Q + W
in this case the cylinder is insulated, so there is no heat transfer
ΔE = W
internal energy can be related to the change in temperature
ΔE = 3/2 n K ΔT
we substitute
3/2 n (T₂-T₁) = W
as the work is on the gas it is negative
W = 3/2 n (T₁- T₂)
Answer:
The new radius of the trajectory of the particle is four times the previous radius
Explanation:
In order to know what is the radius of the trajectory of the charged particle, if its speed is four times as fast, you take into account the following formula, which describes the radius of a charged particle in a magnetic field:
(1)
If the speed of the particle is for time as fast, that is, v' = 4v, you obtain, in the equation (1):
The new radius of the trajectory of the particle is four times the previous radius
