Look first for the relation between deBroglie wavelength (λ) and kinetic energy (K):
K = ½mv²
v = √(2K/m)
λ = h/(mv)
= h/(m√(2K/m))
= h/√(2Km)
So λ is proportional to 1/√K.
in the potential well the potential energy is zero, so completely the electron's energy is in the shape of kinetic energy:
K = 6U₀
Outer the potential well the potential energy is U₀, so
K = 5U₀
(because kinetic and potential energies add up to 6U₀)
Therefore, the ratio of the de Broglie wavelength of the electron in the region x>L (outside the well) to the wavelength for 0<x<L (inside the well) is:
1/√(5U₀) : 1/√(6U₀)
= √6 : √5
Answer:
Velocity of the electron at the centre of the ring, 
Explanation:
<u>Given:</u>
- Linear charge density of the ring=
- Radius of the ring R=0.2 m
- Distance of point from the centre of the ring=x=0.2 m
Total charge of the ring
Potential due the ring at a distance x from the centre of the rings is given by
The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by
Let
be the change in potential Energy given by
Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron
So the electron will be moving with
Answer:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
Explanation:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
Explanation:
∆x=300 m×2
∆t=1.5 s
v=∆x/∆t → v=2×300/1.5 = 400 m/s
Answer
A. the work done on the refrigerant in each cycle is 105kJ
B the coefficient of performance of the refrigerator is 4.8
Explanation
Given data
Work done at high temperature T2 Qh=610kJ
Work done at low temperature T1 Ql=505kJ
We know that the net work done by the refrigerator is expressed as
Wnet= Qh-Ql
=610-505
=105kJ
Also we know that the coefficient of performance is expressed as
COP= Ql/Wnet
COP= 505/105
= 4.8
