An electron with a speed of 1.1 × 107 m/s moves horizontally into a region where a constant vertical force of 3.7 × 10-16 N acts on it. The

Question

An electron with a speed of 1.1 × 107 m/s moves horizontally into a region where a constant vertical force of 3.7 × 10-16 N acts on it. The mass of the electron is 9.11 × 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 22 mm horizontally.

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niczorrrr 1 year 2021-08-31T01:49:54+00:00 1 Answers 4 views 0

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    2021-08-31T01:51:44+00:00

    Explanation:

    GIven data :

    Speed V = 1.1×10⁷ m/s

    Force F = 3.7×10⁻¹⁶N

    Mass of electron = m = 9.11×10⁻³¹Kg

    Horizontal Distance X = 2.2×10⁻²m

    Vertical distance Y = ?

    Solution:

    As we know that,

    F =ma

    where a is the acceleration of electron.

    (3.7×10⁻¹⁶) = (9.11×10⁻³¹) a

    a = 4.06×10¹⁴m/s² upward.

    ∵opposing weight of electron is negligible.

    now we find how long it takes the electron to move 22mm horizontally.

    X =V×T

    T = X/V

    T =2.2×10⁻²/1.1×10⁷

      = 2×10⁻⁹s

    now we can find out the vertical distance Y.

    Y = 0.5aT²

      = (0.5)(4.06×10¹⁴)(2×10⁻⁹)²

      = 8.12×10⁻⁴m

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