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Derive an expression for de Broglie wavelength of electrons. |
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Answer» Solution :(i) An electron of mass m is accelerated through a potential different of V volt. The kinetic energy energy ACQUIRED by the electron is given by `(1)/(2) mv^(2) = eV` (ii) 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(2emV))` (iii) 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)))` ` = ( 12.27 xx 10^(-10))/( sqrt( V )) ` meter ( or ) `lambda = ( 12.27 ) /( sqrt( V )) Å` (iv) 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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