A beam of electrons is accelerated from rest and then passes through a pair of identical thin slits that are 1.35 nm apart. You observe that

Question

A beam of electrons is accelerated from rest and then passes through a pair of identical thin slits that are 1.35 nm apart. You observe that the first double-slit interference dark fringe occurs at ±15.0 ∘ from the original direction of the beam when viewed on a distant screen. Through what potential difference were the electrons accelerated?

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Ngọc Khuê 4 years 2021-08-20T18:50:39+00:00 1 Answers 4 views 0

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  1. RuslanHeatt
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    2021-08-20T18:51:49+00:00
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    Answer:

    The potential difference was 3.09V.

    Explanation:

    The condition for the first dark fringe is

    (1).\: \: \: d\:sin(\theta) = \dfrac{\lambda}{2}

    \lambda = 2d\: sin(\theta).

    putting in numbers we get:

    \lambda = 2 (1.35*10^{-9})(sin(15^o))

    \boxed{\lambda = 6.988*10^{-10}m}

    Now, from de broglie relation we know that

    \lambda = \dfrac{h}{p}

    or

    (2). \: \: p = \dfrac{h}{\lambda }

    where h  is Planck’s constant and p is electron momentum. The momentum is related to kinetic energy by

    E = \dfrac{p^2}{2m},

    this energy we know must come from potential difference; therefore,

    (3). \: \: qV= \dfrac{p^2}{2m}

    substituting the value of p from equation (2), and solving for potential V, we get:

    (4). \: \:V = \dfrac{h^2}{2qm\lambda^2}

    putting in numerical values in equation (4) we get:

    V = \dfrac{(6.63*10^{-34})^2}{2(1.6*10^{-19})(9.1*10^{-31})(6.988*10^{-10})^2}

    \boxed{V = 3.09\:volts}.

    Therefore, the potential difference through which the protons were accelerated was 3.09V.

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