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Show that the de-Broglie's wavelength of electrons accelerated through a potential V volts can be expressed as lamda=(h)/(sqrt(2meV)). What is the de-Broglie wavelength of a 2 kg object moving with a speed of 1 ms^(-1) ? |
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Answer» Solution :Consider an electron of mass m and charge e and initially at REST. If it is accelerated through a potential V volts, then KINETIC energy gained by electron=work done on the electron `therefore (1)/(2)mv^(2)=EV implies v=sqrt((2eV)/(m))` `therefore`de-Broglie wavelength of electron ,br> `lamda=(H)/(p)=(h)/(mv)=(h)/(m*sqrt((2eV)/(m)))=(h)/(sqrt(2meV))` In the numerical problem given we have `m=2kg and v=1ms^(-1)` `therefore lamda=(h)/(mv)=(6.63xx10^(-34))/(2xx1)=3.31xx10^(-34)m`. |
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