The ratio of de-Broglie wavelength of an -particle and proton accelerated from rest by the same potential is 1 m . The value of m is

Step-by-step solution

The de-Broglie wavelength of a particle accelerated through a potential difference V is given by = h 2m₀ qV , where m₀ is the mass and q is the charge of the particle. For a proton: mass m_p , charge q_p = e . For an -particle: mass m_ = 4m_p , charge q_ = 2e . The ratio of their de-Broglie wavelengths is: _ _p = m_p q_p m_ q_ Substituting the values: _ _p = m_p e 4m_p 2e = 1 8 = 1 8 Given that the ratio is 1 m , comparing the two expressions gives m = 8 . Answer: 8