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InterviewSolution
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A proton is accelerated from rest through a potential difference of 500 volts. Find its final momentum. [mp = 1.67 × 10-27 kg, e = 1.6 × 10-19 C] |
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Answer» Data : mp = 1.67 × 10-27 kg, e = 1.6 × 10-19 C, u = 0, V = 500 V Initial KE, KE; = \(\cfrac12\) mpu2 = 0 ∴ ∆KE = KEf – KEi = KEf ∆KE = qV ∴ KEf = qV KEf = \(\cfrac12\) mpv2 = \(\cfrac{p^2_1}{2m_p}\) where pf = mpv ≡ the magnitude of the final momentum of the proton. ∴ Pf2 = 2 mpqV ∴ Pf = \(\sqrt{2m_pqV}\) = \(\sqrt{2(1.67\times10^{-27}\,kg)(1.6\times10^{-19}\,C)(500\,V)}\) = 5.169 × 10-22 kg∙m/s The momentum is directed along the applied electric field. |
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