Answer:
Here\[K=200eV=200\times 1.6\times {{10}^{-19}}J\], \[m=9.1\times {{10}^{-31}}kg,\] \[h=6.6\times {{10}^{-34}}Js\] de-Broglie wavelength of the electron, \[\lambda =\frac{h}{\sqrt{2mK}}\] \[=\frac{6.6\times {{10}^{-34}}}{\sqrt{2\times 9.1\times {{10}^{-31}}\times 200\times 1.6\times {{10}^{-19}}}}\] \[=0.8648\times {{10}^{-10}}m=0.8648\overset{\circ }{\mathop{\text{A}}}\,\] For an electron accelerated through potential difference \[\text{V}\], \[\lambda =\frac{h}{\sqrt{2meV}}\] i.e., \[\lambda \propto \frac{1}{\sqrt{V}}\] \[\therefore \] \[\frac{\lambda '}{\lambda }=\sqrt{\frac{V}{V'}}=\sqrt{\frac{V}{4V}}=\frac{1}{2}\] or \[\lambda '=\frac{\lambda }{2}.\] Thus wavelength would become half of its initial value.
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