Question

A discharge tube contains helium at a low pressure. A large potential difference is applied across the tube. consider a helium atom that has just been ionized due to the detachment of an atomic electron. Find the ratio of the distance travelled by the free electron to that by the positive ion in a short time dt after the ionization.

Answer 1

Let the two particles have charge q mass of electron $m_0=9.1\times {10}^{-31}$ kg, mass of proton $m_p=1.67\times {10}^{-27}kg$ force experienced by Electron = qE
Acceleration $=\frac{qE}{m_0}$
For time dt $S_e=\frac{1}{2}\times \frac{qE}{m_c}\times d{t}^2$
For the positive ion, acceleration
$=\frac{qE}{4\times m_p}{S}_p=\frac{1}{2}\times \frac{qE}{4\times m_p}\times d{t}^2\frac{s_e}{s_p}=\frac{4m_p}{m_e}=\frac{4\times 1.67\times {10}^{-27}}{9.1\times {10}^{-31}}=0.73406\times {10}^{-31}=7340.6$