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Electron should be accelerated which voltage so that its de-Broglie wavelength will become 0.4 Å. |
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Answer» 9410 V `v=(h^(2))/(2melambda^(2))` `=((6.62xx10^(-34))^(2))/(2xx9.1xx10^(-31)xx1.6xx10^(-19)xx(0.4xx10^(-10))^(2))` |
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