Question

A discharge tube contains helium at low pressure. A large potential difference is applied across the tube. Consider a helium atom that has just been ionised 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 ionisation.

Answer 1

Charge on the electron,

- q = −1.6 × 10⁻¹⁹ C

Charge on the helium ion = +q = +1.6 × 10⁻¹⁹ C

Let the electric field in the tube be E.

Mass of electron, m_e = 9.1 × 10⁻³¹ kg

Mass of proton, m_p = 1.67 × 10⁻²⁷ kg

Mass of helium ion, = 4m_p = 4×(1.67 × 10⁻²⁷ ) kg

Magnitude of the force experienced by the electron,

F = qE

Magnitude of acceleration of the electron,

a =\[\frac{qE}{m_e}\]

Distance travelled by the electron in time t,

\[S_e = \frac{1}{2} \times \frac{qE}{m_e} \times t^2.........(1)\]

Acceleration of the positive ion \[= \frac{qE}{4 \times m_p}\]

Distance travelled by the helium ion,

\[S_{He} = \frac{1}{2} \times \frac{qE}{4 \times m_p} \times t^2 ..........(2)\]

The ratio of distance travelled by the given particles is given by

\[\frac{S_e}{S_{He}} = \frac{4 m_p}{m_e}\]

\[\frac{S_e}{S_{He}} = \frac{4 \times 1 . 67 \times {10}^{- 27}}{9 . 1 \times {10}^{- 31}}\]

\[\frac{S_e}{S_{He}} = 0 . 73406 \times {10}^4 = 7340 . 6\]

\[ \Rightarrow \frac{S_e}{S_{He}} = 7340\]