Question

Suppose a diode consists of a cylindrical cathode with a radius of 6.200×10-2 cm, mounted coaxially within a cylindrical anode with a radius of 0.5580 cm. The potential difference between the anode and cathode is 275 V. An electron leaves the surface of the cathode with zero initial speed v(initial) = 0). Find its speed v(final) when it strikes the anode.

 

Answer 1

Given that

radius of the cylindrical cathode, $r_{1}=6.2 \times 10^{-2} \mathrm{~cm}$ radius of the cylindrical anode, $r_{2}=0.5580 \mathrm{~cm}$ potential difference between the anode and cathode, $V=275 \mathrm{~V}$ mass of the electron, $m=9.11 \times 10^{-31} \mathrm{~kg}$ charge on the electron, $q_{e}=1.6 \times 10^{-19} \mathrm{C}$ initial speed of the electron, $v_{\mathrm{i}}=0 \mathrm{~m} / \mathrm{s}$

let the final speed of the electron be $v_{\mathrm{f}}$

The amount of work done on the electron of charge $q_{e}$, when

it is moving between the potential difference $V$ is given by

$$W=q_{\mathrm{e}} V$$

The initial kinetic energy of the electron, $K E_{\mathrm{i}}=\frac{1}{2} m v_{\mathrm{i}}^{2}$ Final kinetic energy of the electron, $K E_{\mathrm{f}}=\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}$ From the work-energy theorem, the work done is equal to the change in kinetic energy.

Hence,

$$\begin{array}{l} W=K E_{\mathrm{f}}-K E_{\mathrm{i}} \\ q_{\mathrm{e}} V=\frac{1}{2} m v_{\mathrm{f}}^{2}-\frac{1}{2} m v_{\mathrm{i}}^{2} \\ q_{\mathrm{e}} V=\frac{1}{2} m v_{\mathrm{f}}^{2}-0 \end{array}$$

Therefore, the speed of the electron at the anode is

$$\begin{aligned} v_{\mathrm{f}}^{2} &=\frac{2 q_{\mathrm{e}} V}{m} \\ v_{\mathrm{f}} &=\sqrt{\frac{2 q_{\mathrm{e}} V}{m}} \\ &=\sqrt{\frac{(2)\left(1.6 \times 10^{-19} \mathrm{C}\right)(275 \mathrm{~V})}{\left(9.11 \times 10^{-31} \mathrm{~kg}\right)}} \\ &=9.82 \times 10^{6} \mathrm{~m} / \mathrm{s} \end{aligned}$$

The speed of the electron at the anode is $9.82 \times 10^{6} \mathrm{~m} / \mathrm{s}$