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)` ?

Explanation

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`.