Modern Physics

JEE Main level, single correct, medium

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An electron is accelerated through a potential difference of V=100 VV = 100\,\text{V}. The de-Broglie wavelength associated with the electron is approximately:

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Answer: (A) 1.227 A˚1.227\,\text{\AA}

λ=12.27V A˚=12.27100=1.227 A˚\lambda = \frac{12.27}{\sqrt{V}}\,\text{\AA} = \frac{12.27}{\sqrt{100}} = 1.227\,\text{\AA}.

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