- Note
- Things to remember

The invisible rays, emerging normally from the cathode of a discharge tube, kept at a pressure of the 0.01 mm of Hg and under very high potential difference of the order of 10-15 KV, supplied from the induction coil, are called cathode rays. These rays are independent of the nature of the gas and their propagation is independent of the position of an anode.

**Properties of Cathode Rays**

- Cathode rays are emitted normally from the surface of the cathode.
- Cathode rays can penetrate the small thickness of matter such as sheets of aluminum foil.
- Cathode rays travel in the straight line and cast sharp shadows of the objects placed in their path.
- Cathode rays carry a negative charge. So they are deflected by electric and magnetic fields.
- Cathode rays carry momentum and kinetic energy.
- They produce heat when they fall upon the matter.
- When cathode rays are suddenly stopped by a target X-rays are produced.
- Cathode rays can ionize the gas through which they pass.
- They affect the photographic plate.
- They travel with 1/10
^{th}the velocity of light. - Cathode rays produce fluorescence when they fall on certain substances like phosphorous.
- Cathode rays can exert mechanical pressures.

Let us suppose that a horizontal beam of electrons, moving with velocity v, passes between two potential plates as shown in a figure. If V be the potential difference between the plates and are separated by a distance d apart, then the field intensity between the plates is given by

$$ E = \frac Vd$$

Therefore, the force on an electron of charge e moving between the plates is given by

$$ F = eE = e \frac Vd $$

This force is directed towards the positive plate.

Since the electric intensity E is vertical, no horizontal force acts on the electron entering the plates. Thus, the horizontal velocity of the beam is unaffected. The vertical acceleration due to gravity does not affect the horizontal motion.

If m_{e} is the mass of an electron, then the vertical acceleration ‘a’ produced in the electron is given by

$$ a= \frac {F}{m_e} = \frac {eE}{m_e} $$

In time t, let an electron describes horizontal distance x and vertical distance y. then,

\begin{align*} x &= vt \dots (i) \\ \text {and} \: y &= ut + \frac 12 at^2 \\ &= 0 + \frac 12 at^2 \\ &= \frac 12 at^2 \\ \text {or,} \: y &= \frac 12 \frac {e}{m_e} . Et^2 \\ &= \frac 12 \frac {e}{m_e}. \frac Vd \left (\frac xv \right )^2 \\ &= \left (\frac 12 \frac {eV}{m_e dv^2} \right ) x^2 \dots (ii) \\ \text {or,} y &= kx^2 \dots (iii) \\ \text {where K} &= \frac {eV}{2m_e v^2 d} \\ \end{align*}

This equation (ii) represents the equation of the parabola. Hence the path of an electron in the electric field is parabolic in nature.

When the electron just passes the plates, x = D. then equation (ii) can be written as

$$ y = \left (\frac 12 \frac {ev}{m_edv^2} \right ) D^2 \dots (iv) $$

The beam then moves on a straight line as shown in the figure.

The time t for which the electron in between the plates is given by

$$ t = \frac Dv $$

Thus, the component of the velocity v_{y} , gained in the direction of the field during time is given by

$$ v_y = \text {acceleration} \times \text {time} = \frac {eV}{m_ed} \times \frac {D}{v} $$

Hence, the angle Ï´ at which the beam emerges from the field is given by

\begin{align*} \tan \theta &= \frac {v_x}{v_y} \\ &= \frac {eV}{m_ed} \times \frac {D}{v} \times \frac 1v \\ &= \frac {eVD}{dm_ev^2} \\ \end{align*}

When v_{x} = v is the horizontal velocity while emerging out of the field.

Let us consider an electron beam moving with speed v horizontally which enters a uniform magnetic field of magnitude B acting perpendicular to the direction of motion as shown in the figure.

As soon as the electron enters into the magnetic field, the force F act on them is given by

$$ F = Be\: v \dots (i) $$

The direction of this force is perpendicular to both B and v, which is given by Fleming’s left-hand rule. Therefore, the speed of the electron remains unaltered but its path is deflected from its original path. Here, the force acting on it still has the same value Bev and since the direction of motion and the magnetic field continue to be mutually perpendicular, the force is perpendicular to the new direction. Therefore, the force only changes the direction of motion but not the speed.

If the field is uniform the force is constant in magnitude and the beam then travels in a circle of radius r. therefore, the force BeV provides the necessary centripetal force.

$$ \text {i.e.} \: Bev = \frac {mv^2}{r} $$

where, m is the mass of an electron.

$$ \therefore \: r = \frac {mv}{Be} = \frac {\text {momentum}}{Be} $$

which gives the radius of the circular path.

Reference

Manu Kumar Khatry, Manoj Kumar Thapa, Bhesha Raj Adhikari, Arjun Kumar Gautam, Parashu Ram Poudel. *Principle of Physics*. Kathmandu: Ayam publication PVT LTD, 2010.

S.K. Gautam, J.M. Pradhan. *A text Book of Physics*. Kathmandu: Surya Publication, 2003.

The invisible rays, emerging normally from the cathode of a discharge tube, kept at a pressure of the 0.01 mm of Hg and under very high potential difference of the order of 10-15 KV, supplied from the induction coil, are called cathode rays.

Cathode rays can penetrate the small thickness of matter such as sheets of aluminum foil and can exert mechanical pressures.

In the motion of electron beam in electric field, if the field is uniform the force is constant in magnitude and the beam then travels in a circle of radius r. therefore, the force BeV provides the necessary centripetal force

.-
## You scored /0

## ASK ANY QUESTION ON Cathode Rays and Motion of Electron Beam in Electric Fields

No discussion on this note yet. Be first to comment on this note