Physics

Physics

Problem

An electron is accelerated from rest through a potential difference of 150 V150\,\text{V}. A proton is to be accelerated so that it acquires the same de Broglie wavelength as this electron. a. (i) Calculate the de Broglie wavelength of the electron after acceleration through 150 V150\,\text{V}. Take me=9.1×10−31 kgm_e = 9.1\times10^{-31}\,\text{kg} and e=1.6×10−19 Ce = 1.6\times10^{-19}\,\text{C}. b. (ii) Determine the potential difference through which the proton must be accelerated to possess the same de Broglie wavelength. Take mp=1.67×10−27 kgm_p = 1.67\times10^{-27}\,\text{kg} and e=1.6×10−19 Ce = 1.6\times10^{-19}\,\text{C}.

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