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An electron is moving with a kinetic energy of `4.55 xx 10^-25 J`. What will be Broglie wavelength for this electron ? |
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Answer» `KE=(1)/(2)mv^(2)=4.55xx10^(-25)` or `(1)/(2)xx9.1xx10^(-31)xxv^(2)=4.55xx10^(-25)` or `v^(2)=(2xx4.55xx10^(-25))/(9.1xx10^(-31))` Applying de Broglie equation, `lamda=(h)/(mv)=(6.6xx10^(34))/(9.1xx10^(-31)xx10^(3))=0.72xx10^(-6)m`. |
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