A proton is accelerated from rest through a potential difference V. After acceleration the mass of the proton is equal to four times its rest mass. Determine the value of V.

For the proton in (a) calculate, after acceleration, its

(i) speed.

(ii) momentum.

\(V = \left[ {\gamma  - 1} \right]938 \times {10^6}\);
γ=4;
V=(3×938×10⁶=)2.81×10⁹ (V) or 2.81 (GV);

or

eV=[γ –1]mc²;
γ=4;
\(V = \left( {\frac{{3 \times 1.67 \times {{10}^{ - 27}} \times 9 \times {{10}^{16}}}}{{1.6 \times {{10}^{ - 19}}}}} \right) = 2.81 \times {10^9}\left( {\rm{V}} \right)\);
Award [3] for a bald correct answer.

(i) recognize that \(4 = \frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\);
to give v=0.968c or 2.90×10⁸(ms^–1);
Allow [2 max] ECF for wrong γ taken from (a).
Award [2] for a bald correct answer.

(ii) \(p = \left( {\gamma {m_0}v = } \right)1.9 \times {10^{ - 18}}\left( {{\rm{kgm}}{{\rm{s}}^{ - 1}}} \right)\) or 3.63×10³(MeVc^-1) or 3.63(GeVc^-1);
Watch for ECF from (a) or (b)(i).