A deuteron and an alpha particl are accelerated with the same potential. Which one of the two has
greater value of de Broglie wavelength associated with it and i

Explanation

de-Broglie `lambda prop (1)/( mv ) `
de-Broglie wavelength of a particle depend upon its mass and charge for same accelerating potential
Mass of a deuteron = `2m_(p)`
Charge of a deuteron `=e `
Mass of an alpha particle `= 4m_(p)`
Charge of an alpha particle `= 2e`
mp-mass of the proton ,e = charge of an electron .
`lambda_(D) //lambda_(a) = m_(a). q_(a) //m_(D).q_(D) = 4m_(p) xx 2e //2m_(p) xxe = 2 lambda_(D) //lambda_(a) = 2 rArr lambda_(D). 2 lambda_(a)`
Thus de Broglie wavelength associated with deuteron is twice of the de Broglie `lambda` of `alpha` particle.