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kondaur [170]
1 year ago
5

A proton, an electron, and a helium nucleus all move at speed v . Rank their de Broglie wavelengths from largest to smallest.

Physics
1 answer:
Varvara68 [4.7K]1 year ago
8 0

Ranking of de Broglie wavelengths from largest to smallest is electron > proton > helium

  • De Broglie proposed that because light has both wave and particle properties, matter exhibits both wave and particle properties. This property has been explained as the dual behavior of matter.
  • From his observations, de Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship

De Broglie's relationship is given by \lambda=\frac{h}{mv}      .....(1) , where λ  is known as de Broglie wavelength and m is mass , v is velocity , h = Plank’s constant.

From equation (1) wavelength and mass has an inverse relation .

Mass of helium is 4 times the mass of the proton and proton has a greater mass than electron.

According to equation (1) , less the mass higher will be the wavelength

Hence electron having less mass have higher wavelength and then proton and then helium having large mass will have less wavelength .

Thus, order should be electron > proton > helium .

Learn about de brogile wavelength more here :

brainly.com/question/16595523

#SPJ4

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1)

When a charge is in motion in a magnetic field, the charge experiences a force of magnitude

F=qvB sin \theta

where here:

For the proton in this problem:

q=1.602\cdot 10^{-19}C is the charge of the proton

v = 300 m/s is the speed of the proton

B = 19 T is the magnetic field

\theta=65^{\circ} is the angle between the directions of v and B

So the force is

F=(1.602\cdot 10^{-19})(300)(19)(sin 65^{\circ})=8.28\cdot 10^{-16} N

2)

The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.

The density of the field lines at any point tells how strong is the magnetic field at that point.

If we observe the field lines around a magnet, we observe that:

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- The density of field lines is lower far from the Poles

Therefore, this means that the magnetic field of a magnet is stronger near the North and South Pole.

3)

The right hand rule gives the direction of the  force experienced by a charged particle moving in a magnetic field.

It can be applied as follows:

- Direction of index finger = direction of motion of the charge

- Direction of middle finger = direction of magnetic field

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In this problem:

- Direction of motion = to the right (index finger)

- Direction of field = downward (middle finger)

- Direction of force = into the screen (thumb)

4)

The radius of a particle moving in a magnetic field is given by:

r=\frac{mv}{qB}

where here we have:

m=6.64\cdot 10^{-22} kg is the mass of the alpha particle

v=2155 m/s is the speed of the alpha particle

q=2\cdot 1.602\cdot 10^{-19}=3.204\cdot 10^{-19}C is the charge of the alpha particle

B = 12.2 T is the strength of the magnetic field

Substituting, we find:

r=\frac{(6.64\cdot 10^{-22})(2155)}{(3.204\cdot 10^{-19})(12.2)}=0.366 m

5)

The cyclotron frequency of a charged particle in circular motion in a magnetic field is:

f=\frac{qB}{2\pi m}

where here:

q=1.602\cdot 10^{-19}C is the charge of the electron

B = 0.0045 T is the strength of the magnetic field

m=9.31\cdot 10^{-31} kg is the mass of the electron

Substituting, we find:

f=\frac{(1.602\cdot 10^{-19})(0.0045)}{2\pi (9.31\cdot 10^{-31})}=1.23\cdot 10^8 Hz

6)

When a charged particle moves in a magnetic field, its path has a helical shape, because it is the composition of two motions:

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2- A circular motion in the direction perpendicular to the magnetic field

The second motion is due to the presence of the magnetic force. However, we know that the direction of the magnetic force depends on the sign of the charge: when the sign of the charge is changed, the direction of the force is reversed.

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7)

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8)

The flux is calculated as

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A=(0.15 m)(0.08 m)=0.012 m^2

So the flux is

\Phi = (5.5)(0.012)(sin 18^{\circ})=0.021 Wb

See the last 7 answers in the attached document.

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