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Question

How is the de-Broglie wavelength associated with an electron accelerated through a potential difference of $$100$$ volts?


Solution

Using energy conservation, $$KE=PE$$ or $$\dfrac{p^2}{2m}=eV$$ or $$p=\sqrt{2meV}$$ where $$p=$$ momentum of electron, $$m=$$ mass of electron, $$e=$$ charge of electron and $$V=$$ potential difference.

Now, de-Broglie wavelength $$\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2meV}}=\dfrac{6.6\times 10^{-34}}{\sqrt{2(9.1\times 10^{-31})(1.6\times 10^{-19})(100)}}=1.22\times 10^{-10} m$$ 

Physics

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