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(a) Show that the de-Broglie wavelength lamda of electrons of kinetic energy K is given by the relation lamda=(h)/(sqrt(2mK)) (b) Calculate the de-Broglie wavelength of an electron beam accelerated through a potential difference of 60 V. |
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Answer» <P> Solution :(a) We know that de-Broglie wavelength of an electron `lamda=(H)/(P)`If K be the kinetic energy and m the mass of electron, then `p=sqrt(2mK)`, and HENCE relation for de-Broglie wavelength becomes: `lamda=(h)/(sqrt(2mK))` (b) Here V=60V and we know that for electron `e=1.6xx10^(-19)C, m=9.11xx10^(-31)kg and h=6.63xx10^(-34)Js` `therefore`de-Broglie wavelength `lamda=(h)/(sqrt(2meV))=(6.63xx10^(-34))/(sqrt(2xx9.11xx10^(-31)xx1.6xx10^(-19)xx60))m` `=1.586xx10^(-10)m or 1.586Å`. |
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