Question

An electron accelarated through 30kV in a cathod ray osaloscope enters a sytem of deflecting plates.The deflecting feilds b/w the plate is 24kV/m the length of the deflecting plate 0.06m and the total deflection produced in the path of electron on the screen is 9mm.What is the distance of the screen from near end of the plate?

Answer 1

An electron accelerated through $30\,\mathrm{kV}$ in a cathode ray oscilloscope enters a system of deflecting plates. The deflecting field between the plates is $24\,\mathrm{kV/m}$, the length of the deflecting plate is $0.06\,\mathrm{m}$ and the total deflection produced in the path of electron on the screen is $9\,\mathrm{mm}$. What is the distance of the screen from near end of the plate?

Solution: Horizontal speed of electron $\nu_h$ before deflecting can be calculated from the law of conservation of energy: $e \cdot V = \frac{m_e \cdot \nu_h^2}{2}$, $\nu_h = \sqrt{\frac{2e \cdot V}{m_e}} = \sqrt{\frac{2 \cdot 1.6 \cdot 10^{-19} \cdot 30 \cdot 10^3}{9.11 \cdot 10^{-31}}} = 1.03 \cdot 10^8 \frac{\mathrm{m}}{\mathrm{s}}$, where $e$ is the charge of electron, $V$ is the accelerating potential difference, $m_e$ is the mass of electron. As you see, according to the laws of classical physics speed of the electron is bigger than $20\%$ of the velocity of light. We must calculate his speed using the formula of relativity theory: $K = m_e \cdot c^2 \cdot (\gamma - 1)$, $\gamma = \left(1 - \nu_h^2 / c^2\right)^{-0.5}$, where $K$ is the kinetic energy of electron and $c$ is the velocity of light.

Then, $K = e \cdot V = 1.6 \cdot 10^{-19} \cdot 30 \cdot 10^3 = 4.8 \cdot 10^{-15}\,\mathrm{J}$; $\gamma = \frac{K}{m_e \cdot c^2} + 1 = \frac{4.8 \cdot 10^{-15}}{9.11 \cdot 10^{-31} \cdot (3 \cdot 10^8)^2} + 1 = 1.0585$;

$$\nu_h = c \cdot \sqrt{1 - 1/\gamma^2} = 3 \cdot 10^8 \cdot \sqrt{1 - 1/1.0585^2} = 9.84 \cdot 10^7 \frac{\mathrm{m}}{\mathrm{s}}.$$

Mass of moving electron can be calculated as $m_e' = \frac{K}{c^2} + m_e = \frac{4.8 \cdot 10^{-15}}{(3 \cdot 10^8)^2} + 9.11 \cdot 10^{-31} = 9.64 \cdot 10^{-31}\,\mathrm{kg}$.

Time of deflecting $t_d$ (time of flying between the plates) of the electron is $t_d = \frac{l}{\nu_h} = \frac{0.06}{9.84 \cdot 10^7} = 6.1 \cdot 10^{-10}\,\mathrm{s}$, where $l$ is the length of the plates.

Vertical acceleration of the moving electron is $a = \frac{F}{m_e'} = \frac{e \cdot E}{m_e'} = \frac{1.6 \cdot 10^{-19} \cdot 24 \cdot 10^3}{9.64 \cdot 10^{-31}} = 4.0 \cdot 10^{15}\,\frac{\mathrm{m}}{\mathrm{s}^2}$, where $F$ is the applied to electron force of deflecting electric field $E$.

Then, vertical speed of electron after deflection will be $\nu_v = a \cdot t_d = 4.0 \cdot 10^{15} \cdot 6.1 \cdot 10^{-10} = 2.44 \cdot 10^6 \frac{\mathrm{m}}{\mathrm{s}}$.

Vertical deflection of electron in the electric field can be calculated as a distance traveled by the uniformly accelerated object: $s_e = \frac{a \cdot t_d^2}{2} = 4.0 \cdot 10^{15} \cdot (6.1 \cdot 10^{-10})^2 / 2 = 7.44 \cdot 10^{-4}\,\mathrm{m} = 0.744\,\mathrm{mm}$.

The rest of the vertical distance electron travels with constant speed $\nu_v$, and then time $t_u$ of the electron uniform motion, which is equal to the time of movement from the near end of the plate to the screen, is:

$$t_u = \frac{d - s_e}{\nu_v} = \frac{9 \cdot 10^{-3} - 7.44 \cdot 10^{-4}}{2.44 \cdot 10^6} = 3.38 \cdot 10^{-9}\,\mathrm{s},$$

where $d$ is the total deflection of the electron.

Electron travels the distance from the near end of the plate to the screen $L$ with constant horizontal velocity. Then, $L = t_u \cdot \nu_h = 3.38 \cdot 10^{-9} \cdot 9.84 \cdot 10^7 = 0.333\,\mathrm{m} = 33.3\,\mathrm{cm}$.

Answer: 33.3 cm.