A deuteron and an alpha particle are accelerated from rest through potential differences V₁ and V₂ respectively. If the de-Broglie wavelength associated with both particles is equal, then the ratio V₁ V₂ is:

Step-by-step solution

The de-Broglie wavelength of a particle accelerated through a potential difference V is given by = h 2mqV , where m is the mass and q is the charge of the particle. Let the mass of a proton be m and its charge be e . For a deuteron: Mass m₁ = 2m Charge q₁ = e Wavelength ₁ = h 2(2m)(e)V₁ = h 4meV₁ For an alpha particle: Mass m₂ = 4m Charge q₂ = 2e Wavelength ₂ = h 2(4m)(2e)V₂ = h 16meV₂ Given that the de-Broglie wavelengths are equal ( ₁ = ₂ ): h 4meV₁ = h 16meV₂ Squaring both sides, we get: 4meV₁ = 16meV₂ V₁ V₂ = 1