Question

The hot filament of the electron gun in a cathode ray tube releases electrons with nearly zero kinetic energy. The electrons are next accelerated under a potential difference of \(5.00 \mathrm{kV}\), before being steered toward the phosphor on the screen of the tube. a) Calculate the kinetic energy acquired by the electron under this accelerating potential difference. b) Is the electron moving at relativistic speed? c) What is the electron's total energy and momentum? (Give both values, relativistic and nonrelativistic, for both quantities.)

Short answer

a) The kinetic energy acquired by the electron under this accelerating potential difference is approximately \(8.00 \times 10^{-16}\,\mathrm{J}\). b) The electron is not moving at a relativistic speed, as its speed (\(1.33 \times 10^{7}\,\mathrm{m/s}\)) compared to the speed of light is significantly less than 1 (\(\frac{v}{c} ≈ 0.044\)). c) The electron's total energy and momentum are as follows: - Relativistic: - Total energy: \(E \approx 8.19 \times 10^{-14}\,\mathrm{J}\) - Momentum: \(p \approx 2.87 \times 10^{-23}\,\mathrm{kg\,m/s}\) - Non-relativistic: - Total energy: \(E = 8.00 \times 10^{-16}\,\mathrm{J}\) - Momentum: \(p \approx 2.82 \times 10^{-23}\,\mathrm{kg\, m/s}\)

Step-by-step solution

Calculate the kinetic energy

To calculate the kinetic energy of the electron, we need to use the expression for kinetic energy in terms of electric potential: \(K_e = e \cdot V\), where \(K_e\) is the kinetic energy, \(e\) is the elementary charge (\(1.60 \times 10^{-19} \mathrm{C}\)), and \(V\) is the potential difference (\(5000 \mathrm{V}\)). Therefore, \(K_e = (1.60 \times 10^{-19}) \cdot (5000)\) \(K_e = 8.00 \times 10^{-16}\,\mathrm{J}\).

Determine if the electron is moving at relativistic speed

To determine if the electron is moving at relativistic speeds, we first find the electron's velocity using the non-relativistic formula for kinetic energy: \(K_e = \frac{1}{2}m_ev^2\), where \(v\) is the velocity of the electron, \(m_e\) is the electron's mass (\(9.11 \times 10^{-31}\,\mathrm{kg}\)). Then we can find, \(v^2 = \frac{2K_e}{m_e}\). \(v = \sqrt{\frac{2(8.00 \times 10^{-16})}{(9.11 \times 10^{-31})}}\) \(v ≈ 1.33 \times 10^{7} \mathrm{m/s}\). We can determine if this speed is relativistic by comparing it to the speed of light (\(c = 3 \times 10^{8} \mathrm{m/s}\)). In our case, \(\frac{v}{c} ≈ 0.044\). Since this value is significantly less than 1, the electron is not moving at a relativistic speed.

Calculate the electron's total energy and momentum

Relativistic: Total energy: \(E = \frac{m_ec^2}{\sqrt{1-\left(\frac{v}{c}\right)^2}}\) Momentum: \(p = \frac{m_ev}{\sqrt{1-\left(\frac{v}{c}\right)^2}}\) Non-relativistic: Total energy: \(E = K_e\) Momentum: \(p = m_ev\) Using the values calculated in Step 1 and Step 2, we have the following values for the electron's total energy and momentum: Relativistic: - Total energy: \(E \approx 8.19 \times 10^{-14}\,\mathrm{J}\) - Momentum: \(p \approx 2.87 \times 10^{-23}\,\mathrm{kg\,m/s}\) Non-relativistic: - Total energy: \(E = 8.00 \times 10^{-16}\,\mathrm{J}\) - Momentum: \(p \approx 2.82 \times 10^{-23}\,\mathrm{kg\, m/s}\)

Key concepts

Relativistic Speed

Electrons moving in a cathode ray tube can achieve high speeds, which raises the question of whether their motion is approaching relativistic levels. In physics, "relativistic" refers to when objects move at speeds considerably close to the speed of light. To determine if an electron's speed is relativistic, we compare its velocity to the speed of light, approximately

• 3.00 \times 10^8 \text{ m/s}

For the electrons in our given problem, after being accelerated through a 5.00 kV potential difference, they achieve a velocity of approximately 1.33 \times 10^7 \text{ m/s}. This value is calculated using the kinetic energy formula and the known mass of an electron. When we examine the ratio of the electron's velocity to the speed of light, we find:

• \( \frac{v}{c} ≈ 0.044 \)

Since this value is much less than 1, the electron’s speed is non-relativistic, meaning relativistic effects like time dilation are negligible here. Understanding whether a speed is relativistic helps determine which formulas to use and how accurate results can be expected.

Kinetic Energy

The kinetic energy of an electron in a cathode ray tube can be calculated from the electric potential energy that accelerates it. The relationship between kinetic energy (\(K_e\)) and electric potential (\(V\)) is: \[K_e = e \cdot V \]where \(e\) is the elementary charge of approximately 1.60 \times 10^{-19} \text{ C}. In our problem, with a 5.00 kV potential difference, the kinetic energy is determined by multiplying the charge with the potential difference:

• \( K_e = (1.60 \times 10^{-19}) \cdot (5000) \)
• \( K_e = 8.00 \times 10^{-16} \text{ J} \)

This value represents the energy acquired by the electron as it's accelerated through the electric field within the tube. It is essential to understand kinetic energy in this context as it directly impacts how fast the electron will be moving and hence, how sharply it can be focused or steered onto the screen.Overall, calculating kinetic energy from potential difference is crucial in applications where high-speed particle beams are used, such as in cathode ray tubes used in older television sets and oscilloscopes.

Momentum Calculation

The momentum of an electron provides insight into its motion and is crucial for understanding its interactions with other particles or fields. Both relativistic and non-relativistic momentum can be calculated, though they differ based on the speed of the electron.For non-relativistic speeds, the momentum (\(p\)) of an electron is given as:\[p = m_e v\]where \(m_e\) is the electron's mass (9.11 \times 10^{-31} \text{ kg}) and \(v\) is the velocity. Using our previous velocity calculation, the non-relativistic momentum becomes:

• \( p \approx 2.82 \times 10^{-23} \text{ kg m/s} \)

Relativistic momentum, however, is more complex due to the speeds approaching light. It requires: \[p = \frac{ m_e v }{ \sqrt{1 - \left(\frac{v}{c}\right)^2} }\]With the velocity calculated, this generates a slightly higher momentum:

• \( p \approx 2.87 \times 10^{-23} \text{ kg m/s} \)

Though in this particular case, due to the electron’s speed not being sufficiently close to light-speed, non-relativistic equations suffice for accuracy. Such momentum calculations are crucial in predicting the electron's trajectory and interactions within fields or matter.