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PolarNik [594]
3 years ago
6

If the particles are oxygen ions carrying a charge of −2e and the magnetic field magnitude is 0.12 T , how long does it take eac

h ion to travel the semicircular path through the magnetic field? The mass of an O2− ion is 2.6 ×10−26kg.
Physics
1 answer:
Natali5045456 [20]3 years ago
4 0

Time taken by each ion to travel the semicircular path through the magnetic field is 34 X 10⁻²⁶s.

<u>Explanation:</u>

Given:

Charge of oxygen ion, q = -2

Magnetic Field, B = 0.12 T

Mass of O₂ , m= 2.6 X 10⁻²⁶ Kg

Time, T = ?

We know,

                T = \frac{2\pi m}{qB}

Therefore,

T = 2 X 3.14 X 2.6 X 10⁻²⁶/ 2 X 0.12

T = 68 X 10⁻²⁶s

For semicircular orbit, Time taken would be T/2

T ₓ = 68 X 10⁻²⁶/2 s

T ₓ = 34 X 10⁻²⁶s

Therefore, time taken by each ion to travel the semicircular path through the magnetic field is 34 X 10⁻²⁶s.

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Given Information:

Mass of electron = m = 9x10⁻³¹ kg

initial speed of electron = v₁ = 0.92c

Force = F =  1.4x10⁻¹³ J

Distance = d = 3 m

Required Information:

Final speed of electron = v₂ = ?

Answer:

Final speed of electron = v₂ = 2.974x10⁸ m/s

Explanation:

As we know from the conservation of energy,

E₂ - E₁ = W

E₂ = E₁ + W

Where E₂ is the final energy of electron and E₁ is the initial energy of electron

The above equation can be written in the form of particle energy

γ₂mc² = γ₁mc² + W

where γ₁ and γ₂ are given by

γ₁ = 1/√1 - (v₁/c)²

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First calculate γ₁

γ₁ = 1/√1 - (0.92c/c)²

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Now calculate γ₂

γ₂ = (γ₁mc² + W)/mc²

First we need to find the work done

W = F*d

W = 1.4x10⁻¹³*3

W = 4.2x10⁻¹³ J

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γ₂ = (2.55*9x10⁻³¹*(3x10⁸)² + 4.2x10⁻¹³)/9x10⁻³¹*(3x10⁸)²

γ₂ = 7.73

Now we can find the new speed of the electron

γ₂ = 1/√1 - (v₂/c)²

Re-arranging the above equation results in

v₂ = c*√(1 - 1/γ₂²)

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a) de Broglie wavelength of a 4-MeV proton: 14.3 fm

b) de Broglie wavelength of a 40-GeV electron: 0.031 fm

Explanation:

a)

The de Broglie wavelength of an object is given by

\lambda=\frac{h}{p} (1)

where

h is the Planck constant

p is the momentum of the particle

Here we want to find the de Broglie wavelength of a 4-MeV proton. The rest of mass of the proton in MeV is

m_0 = 938 MeV

And since 4MeV, this means that the proton is non-relativistic. So its kinetic energy is related to its momentum by

E=\frac{p^2}{2m}

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p=\sqrt{2Em}

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Substituting, we find

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b)

In this case, the electron has kinetic energy of 40 GeV, while the rest mass of an electron is

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Since 40 GeV >> 0.511 MeV, the electron is ultra-relativistic: so we can rewrite its energy as

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The equation (1) can also be rewritten as

\lambda = \frac{hc}{pc}

where c is the speed of light. The quantity at the denominator is the energy, so

\lambda=\frac{hc}{E}

where:

E=40 GeV = 40\cdot 10^9 eV \cdot (1.6\cdot 10^{-19})=6.4\cdot 10^{-9} J is the energy of the electron

And substituting, we find:

\lambda=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{6.4\cdot 10^{-9}}=3.1 \cdot 10^{-17} m = 0.031 fm

Learn more about de Broglie wavelength:

brainly.com/question/7047430

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