A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34×10−27 kgkg and a charge of 1.60×10−19 CC . The deuteron travels in a c

Question

A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34×10−27 kgkg and a charge of 1.60×10−19 CC . The deuteron travels in a circular path with a radius of 6.90 mmmm in a magnetic field with a magnitude of 3.00 TT .

A) Find the speed of the deuteron
B)Find the time required for it to make 12 of a revolution.
C) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

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Dulcie 4 years 2021-07-28T21:30:53+00:00 1 Answers 88 views 0

Answers ( )

  1. kiet
    0
    2021-07-28T21:32:12+00:00
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    Answer:

    a) v=991616m/s

    b) t=2.19*10^-8sec

    c)V=10263V

    Explanation:

    Given that a deuteron of mass 3.34*10^-27kg with a charge of 1.6*10-19C and travel a path of radius 6.90mm which is 0.0069m in a 3 tesla magnetic field

    a)

    We are to find the speed of deuteron

    Recall that F= qVB

    But we know F is also m*a

    Hence

    M*a= qVB

    We’re a is the centripetal acceleration directed inwards

    Ie a=V^2/r

    So M*V^2/r=qVB

    V= qBr/M=1.6*10^-19*3*0.0069/3.34*10^-27

    Velocity (V)=991616.8m/s

    b)

    We are to find the time for 12 revolution

    Note before the path traced by the deuteron is a semi circular path

    Hence T = distance/velocity.

    T= πr/v

    t=3.142*0.0069/991616.8

    t=2.19*10^-8seconds

    c). calculating for the potential difference V

    Remember that the kinetic energy must equal the potential energy

    So

    1/2mv^2=/q/V

    Hence V= 1/2mv^2/q

    V=1/2*3.34*10^-27*991616^2/1.6*10^-19

    V=10263.2V

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