For a proton (mass = 1.673 x 10–27 kg) moving with a velocity of 2.83 x 104 m/s, what is the de Broglie wavelength (in pm)?
1 answer:
Answer:
The value of de Broglie wavelength is 14.0 pm
Explanation:
Given;
mass of proton, m = 1.673 x 10⁻²⁷ kg
velocity of the proton, v = 2.83 x 10⁴ m/s
De Broglie wavelength is given as;
where;
h is planck's constant = 6.626 x 10⁻³⁴ kgm²/s
m is mass of the proton
v is the velocity of the proton
Therefore, the value of de Broglie wavelength is 14.0 pm
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Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find: