Question H3.5: (a) A proton is accelerated from rest by a potential differe...
(a) A proton is accelerated from rest by a potential difference V. Calculate V so that the proton reaches a speed of 0.95 c after acceleration.
(The rest energy of the proton is 938 MeV.)
(b) What accelerating potential is required to accelerate a proton from a speed of 0.95 c to a speed of 0.99 c ?
(a) The gamma factor at a speed of 0.95 c is
\begin{aligned}\gamma &=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \\&=\frac{1}{\sqrt{1-0.95^{2}}} \\&=3.20\end{aligned}The total energy of the proton after acceleration is thus
\begin{aligned}E &=\gamma m_{ 0 } c^{2} \\&=3.20 \times 938 MeV \\&=3004 MeV\end{aligned}From
E=m_{0} c^{2}+q Vwe find
q V=(3004-938) MeV =2066 MeVand so
V=2.1 \times 10^{9} VNotice carefully how we avoid using SI units to make the numerical calculations much easier.
(b) The total energy of the proton at a speed of 0.95 c is (from part (a)) E = 3004 MeV. The total energy at a speed of 0.99 c is (working as in (a)) E = 6649 MeV. The extra energy needed is then 6649 – 3004 = 3645 MeV, and so the accelerating potential must be 3.6 \times 10^{9} V, Notice that a larger potential difference is needed to accelerate the proton from 0.95 c to 0.99 c than from rest to 0.95 c. This is a sign that it is impossible to reach the speed of light. (See also the next example.)