What causes magnetism inside the atom?

Physics

1 answer:

shutvik [7]3 years ago

6 0

Electrons carry a negative electrical charge & produces a magnetic field as they move through space

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A wave with a frequency of 840 cycles per second has what period? Please help me !!

sweet-ann [11.9K]

Answer:

0.005556

Explanation:

The period of a typical class at a high school might be 55 minutes; every 55 minutes a class cycle begins (50 minutes for class and 5 minutes for passing time means that a class begins every 55 minutes). The period for the minute hand on a clock is 3600 seconds (60 minutes); it takes the minute hand 3600 seconds to complete one cycle around the clock.

<u>How to get period from frequency</u>

The formula for the period is T = 1 / f, where "T" is period – the time it takes for one cycle to complete, and "f" is frequency.
To get period from frequency, first convert frequency from Hertz to 1/s.
Now divide 1 by the frequency. The result will be time (period) expressed in seconds.

7 0

3 years ago

A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size

Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of $1.055\times 10^{- 7}$ in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV = $250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J$

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, $m_{p} = 1.67\times 10^{- 27} kg$

The initial size of the wave packet, $\Delta t_{o} = 1 mm = 1\times 10^{- 3} m$

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

$E = m_{p}c^{2}$

$E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J$

This energy of proton is $\simeq 250 GeV$

Thus the speed of the proton, v $\simeq c$

Now, the time taken to cover 1 km = 1000 m of the distance:

T = $\frac{1000}{v}$

T = $\frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s$

Now, in accordance to the dispersion factor;

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}$

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}$