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(C) The angular width of the central maximum in a single-slit diffraction pattern is given by $	heta = \frac{2\lambda}{b}$, where $\lambda$ is the de

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The angular width of the central maximum of the diffraction pattern will decrease.

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(C) The angular width of the central maximum in a single-slit diffraction pattern is given by $	heta = \frac{2\lambda}{b}$, where $\lambda$ is the de Broglie wavelength of the electrons and $b$ is the width of the slit.According to the de Broglie hypothesis, the wavelength is $\lambda = \frac{h}{mv}$, where $h$ is Planck's constant, $m$ is the mass of the electron, and $v$ is its speed.Substituting $\lambda$ into the angular width formula, we get $	heta = \frac{2h}{mbv}$.Since the angular width $	heta$ is inversely proportional to the speed $v$ $(	heta \propto \frac{1}{v})$, increasing the speed $v$ of the electrons will cause the angular width $	heta$ to decrease.Therefore, the correct statement is that the angular width of the central maximum will decrease.

$A$ parallel beam of fast-moving electrons is incident normally on a narrow slit. $A$ screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is $\text{CORRECT}$?

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