Question

A linear charged particle accelerator is driven by an alternating voltage source operating at 10 MHz. Assume that it is used to accelerate electrons. After a few drift-tubes, the electrons attain a velocity \( 2.9 \times 10^8 \, \text{m/s} \). The minimum length of each drift-tube, in m, to accelerate the electrons further (rounded off to one decimal place) is \(\underline{\hspace{2cm}}\).

Hint:

The minimum length of the drift-tube is the distance traveled by the electron in one period of the alternating voltage.

Answer 1

Correct Answer: 14

The velocity of the electrons is given as \( v = 2.9 \times 10^8 \, \text{m/s} \). The frequency of the alternating voltage source is \( f = 10 \, \text{MHz} = 10^7 \, \text{Hz} \). The time period \( T \) of the alternating voltage is: \[ T = \frac{1}{f} = \frac{1}{10^7} = 10^{-7} \, \text{s}. \] The minimum length of the drift-tube corresponds to the distance the electron travels in one period of the alternating voltage: \[ L = v \cdot T = (2.9 \times 10^8) \cdot (10^{-7}) = 29 \, \text{m}. \] Thus, the minimum length of each drift-tube is 29 meters, and rounded to one decimal place, it is 14.0 meters.