Electrons are accelerated to a speed of 4.0 × 10⁴m/s and are aimed at a double-slit apparatus.Interference fringes are detected.If the electrons were replaced by neutrons,what speed must neutrons have to produce interference fringes with the same fringe spacing as that observed with the electrons? A)2.2 × 10¹ m/s B)7.3 × 10⁷ m/s C)1.7 × 10⁶ m/s D)9.3 × 10² m/s

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Quizplus · Accepted Answer

The fringe spacing in a double-slit experiment is given by: $$\Delta y = \frac{\lambda L}{d}$$ where $\lambda$ is the wavelength of the particle, $L$ is the distance between the double-slit apparatus and the screen, and $d$ is the distance between the two slits. For electrons, we have: $$\Delta y_e = \frac{hL}{md}$$ where $m$ is the mass of the electron. We can equate this to the fringe spacing for neutrons: $$\Delta y_n = \frac{hL}{md_n}$$ where $d_n$ is the distance between the two slits for neutrons. Since the fringe spacing is the same for both electrons and neutrons, we have: $$\Delta y_e = \Delta y_n$$ $$\frac{hL}{md_e} = \frac{hL}{md_n}$$ $$\frac{1}{d_e} = \frac{1}{d_n} \frac{m}{m_e}$$ where $m_e$ is the mass of the electron. We can solve for $d_n$: $$d_n = d_e \frac{m}{m_e}$$ The speed of a particle is related to its wavelength by the de Broglie relation: $$\lambda = \frac{h}{p} = \frac{h}{mv}$$ where $p$ is the momentum of the particle and $v$ is its speed. Therefore, the wavelength of the neutrons must be the same as that of the electrons in order to produce the same fringe spacing. We have: $$\lambda_n = \lambda_e$$ $$\frac{h}{mv_n} = \frac{h}{mv_e}$$ $$v_n = v_e \frac{m_e}{m}$$ Substituting the given values, we get: $$v_n = 4.0 imes 10^{14} ext{ m/s} imes \frac{9.11 imes 10^{-31} ext{ kg}}{1.675 imes 10^{-27} ext{ kg}} = 2.2 imes 10^{11} ext{ m/s}$$ Therefore, the answer is (A) 2.2 × 10¹ m/s.

Electrons Are Accelerated to a Speed of 4