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15 November, 02:29

A proton (? p = + ?, mp = 1.0 u; where u = unified mass unit ≃ 1.66 * 10-27kg), a deuteron (? = + ?, m? = 2.0 u) and an alpha particle (? = + 2?, m? = 4.0 u) are accelerated from rest through the same potential difference?, and then enter the same region of uniform magnetic field?⃗⃗, moving perpendicularly to the direction of the magnetic field.

a. What is the ratio of the proton's kinetic energy Kp to the alpha particle's kinetic energy K?

b. What is the ratio of the deuteron's kinetic energy K? to the alpha particle's kinetic energy K?

c. If the radius of the proton's circular orbit? p = 10 cm, what is the radius of the deuteron's orbit?

d. What is the radius of the alpha particle's orbit?

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  1. 15 November, 02:49
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    a. 1/2 b. 1/2 c, 20 cm d. 40 cm

    Explanation:

    Here is the complete question

    A proton ( = +, = 1.0 u; where u = unified mass unit ≃ 1.66 * 10-27kg), a deuteron ( = +, = 2.0 u) and an alpha particle ( = + 2, = 4.0 u) are accelerated from rest through the same potential difference, and then enter the same region of uniform magnetic field ⃗⃗, moving perpendicularly to the direction of the magnetic field.

    A) What is the ratio of the proton's kinetic energy to the alpha particle's kinetic energy?

    B) What is the ratio of the deuteron's kinetic energy to the alpha particle's kinetic energy?

    C) If the radius of the proton's circular orbit = 10 cm, what is the radius of the deuteron's orbit?

    D) What is the radius of the alpha particle's orbit?

    Solution

    a. For both particles, kinetic energy = electric potential energy

    For proton K. E = K₁ = 1/2m₁v₁² = + eV, for alpha particle K. E = K₂ = 1/2m₂v₂² = + 2eV

    where m₁, m₂ and v₁, v₂ are the respective masses and velocities of the proton and alpha particle. So, the ratio of their kinetic energies is

    1/2m₁v₁²/1/2m₂v₂² = + eV/+2eV

    m₁v₁²/m₂v₂² = 1/2.

    So the ratio K₁/K₂ = 1/2

    b. For both particles, kinetic energy = electric potential energy

    For deuteron K₁ = 1/2m₁v₁² = + eV, for alpha particle K₂ = 1/2m₂v₂² = + 2eV

    where m₁, m₂ and v₁, v₂ are the respective masses and velocities of the deuteron and alpha particle. So, the ratio of their kinetic energies is

    1/2m₁v₁²/1/2m₂v₂² = + eV/+2eV

    m₁v₁²/m₂v₂² = 1/2.

    So the ratio K₁/K₂ = 1/2

    c. The radius of the proton's circular is gotten from the centripetal force which equal the magnetic force. So,

    mv²/r = Bev

    r₁ = mv/Be

    Since mass of deuteron m₂ equals twice mass of proton m₁, m₂ = 2m₁

    So, radius of deuteron's circular orbit equals

    r₂ = m₂v/Be = 2m₁v/Be = 2r₁ = 2 * 10 cm = 20 cm

    d. The radius of the alpha particle is given by r₃ = m₃v/Be. Since mass of alpha particle equal four times mass of proton, m₃ = 4m₁.

    So, radius of alpha particle's circular orbit equals

    r₃ = m₃v/Be = 4m₁v/Be = 4r₁ = 4 * 10 cm = 40 cm
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