If ₀ is the de-Broglie wavelength for a proton accelerated through a potential difference of 100 ~V , the de-Broglie wavelength for -particle accelerated through the same potential difference is

Step-by-step solution

V=100 volt Given, ₀= de-Broglie wavelength of proton ₀= h 2 m_p q_p V de-Broglie wavelength of -particle aligned & = h 2 m_ q_ V m_ & =4 m_p aligned (Mass of -particle is four times of proton) q_ =2 q_p (Charge of -particle is twice of proton) aligned & = h 2 (4 m_p ) (2 q_p ) V & = 1 8 h 2 m_p q_p V = ₀ 8 = ₀ 2 2 aligned