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 contains 2 protons and 2 neutrons. Therefore, the mass M of the alphaa particle is 4 times that of a proton (m,) (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(2MqV) = h/sqrt(2 xx (4m_p) xx (2e) V)`
`= (6.62 xx 10^(-34))/sqrt(2 xx 4xx1.67xx 10^(-27)xx 2 xx1.6 xx 10^(-19)xx 400)`
`(6.626 xx 10^(-34))/(4xx 20 xx 10^(-23)sqrt(2xx 1.6xx 10^(-19)xx 400))`