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Derive an expression for De Broglie wavelength. |
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Answer» SOLUTION :An electron of mass m is accelerated through a potential difference of V volt. The kinetic energy ACQUIRED by the electron is given by `(1)/(2) mv^(2) = EV` Therefore, the speed v of the electron is `v = sqrt((2eV)/(m))` Hence, the de Broglie wavelength of the electron is `lambda = (h)/(mv) = (h)/(sqrt(2 EMV))` Substituting the known values in the above equation, we get `lambda = (6.626 xx 10^(-34))/(sqrt(2V xx 1.6 xx 10^(-19) xx 9.11 xx 10^(-31)))` `lambda = (12.27 xx 10^(-10))/(sqrt(V))"meter" (or) lambda = (12.27)/(sqrt(V)) Å` For example, if an electron is accelerated through a potential difference of 100 V, then its de Broglie wavelength is 1.227 `Å`. Since the kinetic energy of the electron, K = eV, then the de Broglie wavelength associated with electron can be also written as `lambda = (h)/(sqrt(2mK))` |
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