The ratio of de Broglie wavelength of a deutron with kinetic energy E to that of an alpha particle with kinetic energy 2 E , is n: 1 . The value of n is _ _ _ _ . (Assume mass of proton = mass of neutron) :

Step-by-step solution

The de Broglie wavelength is = h 2mE . For a deuteron (mass m_d = 2m_p ) with energy E : _d = h 2 2m_p E = h 4m_p E For an alpha particle (mass m_ = 4m_p ) with energy 2E : _ = h 2 4m_p 2E = h 16m_p E The ratio: _d _ = 16m_p E 4m_p E = 4 = 2