Q2PE. - College Physics (Urone)

Question

An evacuated tube uses an accelerating voltage of \(40{\rm{ }}kV\) to accelerate electrons to hit a copper plate and produce \(x\) rays. Non-relativistically, what would be the maximum speed of these electrons?

Step-by-Step Solution

Answer

The maximum speed of the electrons are \({v_e} = 1.19 \times {10^8}\;m/s\).

: Given Information

• Accelerating voltage of evacuated tube – \(40{\rm{ }}kV\)

Concept Introduction

The conservation of energy principle states that the sum of a system's initial energies plus any external forces' work on the system equals the system's final energies.

\({E_i} + W = {E_f}...................(1)\)

Electric Potential Energy: If a particle with charge \(q\) is positioned at a location where a charged object's electric potential is \(V\), the particle-object system's electric potential energy is -

\(PE = qV.......................(2)\)

An object's kinetic energy is described as -

\(KE = \frac{1}{2}m{v^2}...................(3)\)

where \(v\) is the speed with relation to the selected coordinate system and \(m\) is the object's mass. You should be aware that kinetic energy is always positive.

Deducing kinetic and potential energy

Identify the system as the electron and the electric field in the evacuated tube. We choose the zero level of the \(V\) field to be at the initial position of the electron. Since the electron's mass is so little compared to its charge that the outcome would never approach becoming significant, disregard the potential energy due to gravity. Furthermore, the flight takes only a brief period.

The system is initially devoid of both kinetic energy and electric potential energy since the electric potential is zero (the electron is at rest). Positive kinetic energy and a negative electric potential energy must be present in the system's ultimate state. Using Equation \((2)\), the electric potential energy is defined as \(PE = e\;V\) since the electron's charge is negative and the electric potential in the final configuration is positive. Using Equation \((3)\), determine the electron's kinetic energy \(KE = \frac{1}{2}{m_e}v_e^2\).

Using the energy principle

The conservation of energy principle from Equation \((1)\) tells us, that since there are no nonconservative the system in the final state. So,

\(0 = PE + KE\\0 =  - eV + \frac{1}{2}{m_e}v_e^2\\\frac{1}{2}{m_e}v_e^2 = eV\ end\)

Solving for \({v_e}\), it is obtained:

\({m_e}v_e^2 = 2eV\\v_e^2 = \frac{{2eV}}{{{m_e}}}\\{v_e} = \sqrt {\frac{{2eV}}{{{m_e}}}} \ end\)

Calculating the speed

Entering the values for \(e\), \(V\), and \({m_e}\), gives –

\({v_e} = \sqrt {\frac{{2\left( {1.60 \times {{10}^{ - 19}}{\rm{ }}C} \right)(40{\rm{ }}kV)\left( {\frac{{1000\;V}}{{1{\rm{ }}kV}}} \right)}}{{9.11 \times {{10}^{ - 31}}\;kg}}} \\ = 1.19 \times {10^8}\;m/s\ end\)

 

Therefore, the speed is obtained as \({v_e} = 1.19 \times {10^8}\;m/s\).