Calculate the de-Brogile wavelength of an electron.

Explanation

Mass of electron `=m_(e) = 9.1 xx 10^(-31)kg`
When electron is accelerated through a potential difference V, then energy acquired by it is given by E = eV
where e= Charge on an electron
But `E= (1)/(2) mv^(2)= ((mv)^(2))/(2m)`
`rArr mv= sqrt(2mE)` ...(i)
where v= velocity of electron
Now de-Broglie wavelength of electron `=lamda= (h)/(mv)`
`lamda= (h)/(sqrt(2mE))` [Using (i)] ...(ii)
Put value of E in eqn (ii) `lamda= (h)/(sqrt(2me V))`
Now `h= 6.62 xx 10^(-34) Js, e= 1.6 xx 10^(-19) C`
`lamda= (6.62 xx 10^(-34))/(sqrt(2 xx 9.1 xx 10^(-31) xx 1.6 xx 10^(-19) xx V))`
`lamda= (12.27 xx 10^(-10))/(sqrtV) m`
`lamda =(12.27)/(sqrtV) Å`