question_answer

Calculate the de-Broglie wavelength associated with an electron of energy\[\mathbf{200}\text{ }\mathbf{eV}\]. What will be the change in this wavelength if the accelerating potential is increased to four times its earlier value?

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.