A proton and an `alpha`particle are accelerated through the same potential difference. Which one of the two has
greater de Broglie wavelength,

Explanation

The de Broglie wavelengt is related to the accelerating potential as
`lamda = (h)/(sqrt(2meV))`
`:. lamda_(p) = (h)/(sqrt(2m_(p)e_(p)V)) and , lamda_(alpha) = (h)/(sqrt(2m_(alpha) e_(alpha) V))`
`:. (lamda_(p))/(lamda_(alpha)) = (h)/(sqrt(2m_(p) e_(p) V)) xx (sqrt(2m_(alpha) e_(alpha) V))/(h) = (sqrt(m_(alpha) e_(alpha)))/(sqrt(m_(p) e_(p)))`
Now, mass and charge of `alpha-`particle is greater than that of proton.
`:. lamda_(p) gt lamda_(alpha)` for the same potential difference