Answer:
the magnetic field experienced by the electron is 0.0511 T
Explanation:
Given the data in the question;
Wavelength λ = 21 cm = 0.21 m
we know that Bohr magneton μ
is 9.27 × 10⁻²⁴ J/T
Plank's constant h is 6.626 × 10⁻³⁴ J.s
speed of light c = 3 × 10⁸ m/s
protein spin causes magnetic field in hydrogen atom.
so
Initial potential energy = -μB × cos0°
= -μB × 1
= -μB
Final potential energy = -μB × cos180°
= -μB × -1
= μB
so change in energy will be;
ΔE = μB - ( -μB )
ΔE = 2μB
now, difference in energy levels will be;
ΔE = hc/λ
2μB = hc/λ
2μBλ = hc
B = hc / 2μλ
so we substitute
B = [(6.626 × 10⁻³⁴) × (3 × 10⁸)] / [2(9.27 × 10⁻²⁴) × 0.21 ]
B = [ 1.9878 × 10⁻²⁵ ] / [ 3.8934 × 10⁻²⁴ ]
B = 510556326.09
B = 0.0511 T
Therefore, the magnetic field experienced by the electron is 0.0511 T
Much of the precipitation in large bodies of water occurs at the surface. The ocean loses about 37000 km cubed considering evaporation and precipitation.
Answer:
The separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)
Explanation:
The relationship between energy and wavelength is expressed below:
E = hc/λ
λ = hc/EK - EL
Considering the condition of Bragg's law:
2dsinθ = mλ
For the first order Bragg's law of reflection:
2dsinθ = (1)λ
2dsinθ = hc/EK - EL
d = hc/2sinθ(EK - EL)
Where 'd' is the separation distance between the parallel planes of an atom, 'h' is the Planck's constant, 'c' is the velocity of light, θ is the angle of reflection, 'EK' is the energy of the K shell and 'EL' is the energy of the K shell.
Therefore, the separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)
<span> For any body to move in a circle it requires the centripetal force (mv^2)/r.
In this case a ball is moving in a vertical circle swung by a mass less cord.
At the top of its arc if we draw its free body diagram and equate the forces in radial
direction to the centripetal force we get it as T +mg =(mv^2)/r
T is tension in cord
m is mass of ball
r is length of cord (radius of the vertical circle)
To get the minimum value of velocity the LHS should be minimum. This is possible when T = 0. So
minimum speed of ball v at top =sqrtr(rg)=sqrt(1.1*9.81) = 3.285 m/s
In the second case the speed of ball at top = (2*3.285) =6.57 m/s
Let us take the lowest point of the vertical circle as reference for potential energy and apllying the conservation of energy equation between top & bottom
we get velocity at bottom as 9.3m/s.
Now by drawing the free body diagram of the ball at the bottom and equating the net radial force to the centripetal force
T-mg=(mv^2)/r
We get tension in cord T=13.27 N</span>
Answer: a) the force will be repulsive
b) the ratio of the new force to the old force will be 2
c) O
Explanation:
a) since charge -Q is moved from A to B, this implies that sphere A is negatively charged. The two spheres are now negatively charged and will repel themselves.
b) initial force will be -q(-Q)/d2
Adding extra charge -Q will cause change on B to become -2Q
The new force will be - 2Q(-q)/d2
Dividing new force by old force will give 2
C) if B is neutralized, the net charge becomes 0 and there will be no force on it.
