Find the de Broglie wavelength associated with an alpha particle which is accelerated through a potential difference of 400 V. Given that the mass of the proton is `1.67 xx 10^(-27)` kg.

Explanation

An alpha particle cpmtaoms 2 protons and 2 neutrons. Therefore, the mass M of the alpha particle is 4 times that of a proton `(m_(p))` (or a neutron) and its charge q is twice that of a proton (+e).
The de Broglie wavelength associated with it is
`lambda = (h)/(sqrt(2 MqV)) = (h)/(sqrt(2 xx (4m_(p)) xx (2e) V))`
`= (6.626 xx 10^(-34))/(sqrt(2 xx 4 xx 1.67 xx 10^(27) xx 2 xx 1.6 xx 10^(-19) xx 400))`
`= (6.626 xx 10^(-34))/(4 xx 20 xx 10^(-23) sqrt(1.67 xx 1.6)) = 0.00507 Å`