The proton and - particle are accelerated through same potential difference. Then the ratio of the de-Broglie wavelength of proton and - particle is (mass of -particle is 4 times mass of proton, charge of -particle is 2 times charge of proton)

Step-by-step solution

The de-Broglie wavelength of a particle accelerated through a potential difference V is given by: = h 2mqV For a proton: _p = h 2m_p q_p V For an -particle: _ = h 2m_ q_ V The ratio of their de-Broglie wavelengths is: _p _ = m_ q_ m_p q_p Given m_ = 4m_p and q_ = 2q_p , substituting these values: _p _ = 4m_p 2q_p m_p q_p = 8 = 2 2 Answer: 2 2