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 \(9.82 \times 10^{6} \mathrm{~m} / \mathrm{s}\)