Question

Through what potential difference should an electron be accelerated to give it a speed of (a) 0.6c, (b) 0.9c and (c) 0.99c?

Answer 1

Kinetic energy of electron = mc² − m₀c²   ..........(1)
Suppose the electron is accelerated through a potential difference of V. Then,

KE of electron = eV

\[m = \frac{m_0 c^2}{\sqrt{1 - v^2 / c^2}}\]

Putting the values of m and KE in eq. (1), we get

\[eV = \frac{m_0 c^2}{\sqrt{1 - v^2 / c^2}} - m_0 c^2............(2)\]

(a) Velocity of electron, v = 0.6c

Rest mass of electron, m₀ = \[9 . 1 \times {10}^{- 31} kg\]

Charge of electron, e = \[1 . 6 \times {10}^{- 19} C\]

Putting the values of m₀, v and e in eq. (2), we get

\[eV = \frac{9 . 1 \times {10}^{- 31} \times 9 \times {10}^{16}}{\sqrt{1 - \frac{0 . 36 c^2}{c^2}}} - 9 . 1 \times {10}^{- 31} \times 9 \times {10}^{16} \]
\[ \Rightarrow V = \frac{9 . 1 \times {10}^{- 31} \times 9 \times {10}^{16} \left( 1 . 25 - 1 \right)}{1 . 6 \times {10}^{- 19}}\]
\[ \Rightarrow V = 0 . 51 \times 0 . 25 MV\]
\[ \Rightarrow V = 0 . 1275 Mev = 127 . 5 kV\]

(b) Putting v = 0.9c in eq. (2), we get
V = 661 kV

(c) Putting v = 0.99c in eq. (2), we get
V = 3.1 MV