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 an ode, \(r_{2}=0.5580 \mathrm{~cm}\)

potential difference between the an ode and cathode, \(V=275 \mathrm{~V}\)

mass of the electron, \(m=9.11 \times 10^{-31} \mathrm{~kg}\)

charge on the electron, \(q_{\mathrm{e}}=1.6 \times 10^{-19} \mathrm{C}\)

initial speed of the electron, \(v_{i}=0 \mathrm{~m} / \mathrm{s}\)

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

$$ W=q_{e} V $$

Initial kinetic energy of the electron, \(K E_{i}=\frac{1}{2} m v_{i}^{2}\)

Final kinetic energy of the electron, \(K E_{\mathrm{f}}=\frac{1}{2} m v_{\mathrm{f}}^{2}\) From the work -energy the orem, the work done is equal to

the change in kinetic energy.

Hence,

$$ W=K E_{\mathrm{f}}-K E_{\mathrm{i}} $$

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

Therefore, the speed of the electron at the anode is

$$ v_{\mathrm{f}}^{2}=\frac{2 q_{\mathrm{e}} V}{m} $$

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