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(B) Accelerated electrons exhibit wave nature and therefore they form an interference pattern. The fringe width $\beta$ of the pattern is given by $\b

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is half the original fringe width

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(B) Accelerated electrons exhibit wave nature and therefore they form an interference pattern. The fringe width $\beta$ of the pattern is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the de Broglie wavelength associated with the electrons.The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2meU}}$.Substituting this into the fringe width formula,we get $\beta = \frac{hD}{d\sqrt{2meU}}$.Since $h, D, d, m,$ and $e$ are constants,we have $\beta \propto \frac{1}{\sqrt{U}}$.Let $\beta_i$ be the initial fringe width with potential $U$ and $\beta_f$ be the final fringe width with potential $4U$.Then,$\frac{\beta_f}{\beta_i} = \sqrt{\frac{U}{4U}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,$\beta_f = \frac{1}{2} \beta_i$,which means the fringe width becomes half of its initial value.

$A$ beam of monoenergetic electrons,which have been accelerated from rest by a potential $U$,is used to form an interference pattern in a Young's double slit experiment. The electrons are now accelerated by potential $4U$. Then,the fringe width

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