Question

A hot filament liberates an electron with zero initial velocity. The anode potential is $1200\,V$. The speed of the electron when it strikes the anode is

• A. $1.5 \times 10^5 \,ms^{-1}$
• B. $2.5 \times 10^6 \,ms^{-1}$
• C. $2.1 \times 10^7 \,ms^{-1}$
• D. $2.5 \times 10^8 \,ms^{-1}$

Answer 1

The Correct Option is C

The energy gained by the electron when accelerated through a potential \(V\) is given by: \[ E = eV \] Where \(e\) is the charge of the electron and \(V\) is the potential. The kinetic energy of the electron when it strikes the anode is: \[ E = \frac{1}{2} m v^2 \] Equating the two expressions for energy: \[ eV = \frac{1}{2} m v^2 \] Solving for \(v\) (speed of the electron): \[ v = \sqrt{\frac{2eV}{m}} \] Substituting the values for the charge of the electron \(e = 1.6 \times 10^{-19} \, \text{C}\), the mass of the electron \(m = 9.1 \times 10^{-31} \, \text{kg}\), and \(V = 1200 \, \text{V}\): \[ v = \sqrt{\frac{2 \times (1.6 \times 10^{-19}) \times 1200}{9.1 \times 10^{-31}}} \] This simplifies to: \[ v \approx 2.1 \times 10^7 \, \text{ms}^{-1} \] 

Answer 2

The electron is liberated from the filament and is accelerated through a potential difference of 1200 V. The work done on the electron is equal to the kinetic energy acquired by the electron. The energy gained by the electron is given by: \[ \text{Kinetic Energy} = eV \] where:
\( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \, \text{C} \)),
\( V \) is the potential difference (\( 1200 \, \text{V} \)).

This energy is also expressed as: \[ \text{Kinetic Energy} = \frac{1}{2} m v^2 \] where: - \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \, \text{kg} \)), - \( v \) is the velocity of the electron. Equating the two expressions for kinetic energy: \[ eV = \frac{1}{2} m v^2 \] Solving for \( v \): \[ v = \sqrt{\frac{2 e V}{m}} \] Substitute the known values: \[ v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 1200}{9.11 \times 10^{-31}}} \] \[ v = \sqrt{\frac{3.84 \times 10^{-16}}{9.11 \times 10^{-31}}} \] \[ v = \sqrt{4.22 \times 10^{14}} = 2.1 \times 10^7 \, \text{ms}^{-1} \]

Thus, the speed of the electron when it strikes the anode is \({2.1 \times 10^7 \, \text{ms}^{-1}} \).