A deuteron and an alpha particle are accelerated with the same accelerating potential. Which one of the two has 1) greater value of de-Broglie wavelength associated with it and 2) less kinetic energy?

Using conservation of energy, change in electrostatic energy equals the kinetic energy gained. E = 1/2mv² = qV p = mv = √(2mqV)

1. De-broglie wavelength is given by: λ = h/p λ = h/√(2mqV)
   λd/λα = √(mαqα/mdqd)
   λd/λα = √(4 x 2/1 x 2) = 2
   Hence, wavelength of deuteron is more than the wavelength of the alpha particle.
2. Kinetic energy, E = qV as discussed above.
   Ed/Eα = qd/qα = 1/2
   Hence, kinetic energy of alpha particle is less than the kinetic energy of deuteron.