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(B) When electrons are accelerated through a potential of $V$ volts,the momentum $p$ gained by the electrons is $p = \sqrt{2meV}$,where $m$ is the mas

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$0.7$

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(B) When electrons are accelerated through a potential of $V$ volts,the momentum $p$ gained by the electrons is $p = \sqrt{2meV}$,where $m$ is the mass of the electron,$e$ is the charge,and $V$ is the potential.The de Broglie wavelength $\lambda$ associated with these electrons is $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$.The fringe width $\beta$ in a Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $D$ is the distance to the screen and $d$ is the slit separation.Substituting $\lambda$,we get $\beta = \frac{hD}{d\sqrt{2meV}}$. Thus,$\beta \propto \frac{1}{\sqrt{V}}$.If the potential is doubled $(V_f = 2V_i)$,the new fringe width $\beta_f$ is $\beta_f = \frac{\beta_i}{\sqrt{2}} = \frac{\omega}{\sqrt{2}} \approx 0.707\omega$.Therefore,the width is close to $0.7\omega$.

Electrons accelerated from rest by an electrostatic potential are collimated and sent through a Young's double slit experiment. The fringe width is $\omega$. If the accelerating potential is doubled,then the width is now close to ............. $\omega$.

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