An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an electric field.

Question

An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an electric field. The particle then begins to move. Find the speed of the alpha particle after it has moved through a potential difference of â3.45Ã10^â3 V .The charge and the mass of an alpha particle are qα = 3.20Ã10^â19 C and mα = 6.68Ã10â27 kg , respectively.

Mechanical energy is conserved in the presence of which of the following types of forces?

a. electrostatic
b. frictional
c. magnetic
d. gravitational

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Orla Orla 3 years 2021-09-04T08:28:22+00:00 1 Answers 14 views 0

Answers ( )

  1. Xavia
    0
    2021-09-04T08:30:07+00:00
    Reply

    Answer:

    Speed = 575 m/s

    Mechanical energy is conserved in electrostatic, magnetic and gravitational forces.

    Explanation:

    Given :

    Potential difference, U = $-3.45 \times 10^{-3} \ V$

    Mass of the alpha particle, $m_{\alpha} = 6.68 \times 10^{-27} \ kg$

    Charge of the alpha particle is, $q_{\alpha} = 3.20 \times 10^{-19} \ C$

    So the potential difference for the alpha particle when it is accelerated through the potential difference is

    $U=\Delta Vq_{\alpha}$

    And the kinetic energy gained by the alpha particle is

    $K.E. =\frac{1}{2}m_{\alpha}v_{\alpha}^2 $

    From the law of conservation of energy, we get

    $K.E. = U$

    $\frac{1}{2}m_{\alpha}v_{\alpha}^2 = \Delta V q_{\alpha}$

    $v_{\alpha} = \sqrt{\frac{2 \Delta V q_{\alpha}}{m_{\alpha}}}$

    $v_{\alpha} = \sqrt{\frac{2(3.45 \times 10^{-3 })(3.2 \times 10^{-19})}{6.68 \times 10^{-27}}}$

    $v_{\alpha} \approx 575 \ m/s$

    The mechanical energy is conserved in the presence of the following conservative forces :

    — electrostatic forces

    — magnetic forces

    — gravitational forces

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