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 an electron under this accelerating potential difference. b) Is the electron moving at relativistic speed? c) What are the electron's total energy and momentum? (Give both values, relativistic and nonrelativistic, for both quantities.)

Short answer

In summary, under the given potential difference of 5.0 kV, the electron acquires a kinetic energy of 8.00 x 10^(-16) J. As the electron's speed is much smaller than 10% of the speed of light, it is not traveling at a relativistic speed. However, the relativistic and nonrelativistic values for the total energy (8.19 x 10^(-14) J) and momentum (2.71 x 10^(-24) kg m/s) of the electron do not differ significantly in this case.

Step-by-step solution

(a) Calculate the kinetic energy acquired by an electron

To calculate the kinetic energy acquired by an electron under the given potential difference, we can use the following formula: \(K.E=(e)(V)\) where \(K.E\) is the kinetic energy, \(e = 1.60 × 10^{-19} C\) is the charge of the electron, and \(V = 5.00 kV\) is the potential difference. Plug the values: \(K.E =(1.60 × 10^{-19} C)(5.00 × 10^{3} V)\) \(K.E = 8.00 × 10^{-16} J\) Therefore, the kinetic energy acquired by an electron under this accelerating potential difference is \(8.00 × 10^{-16} J\).

(b) Is the electron moving at relativistic speed?

To determine if an electron is moving at relativistic speed, we can compare its speed with a significant fraction of the speed of light (c), for instance, about 10%: \(K.E = \frac{1}{2}m_e v^2\) where \(m_e = 9.11 × 10^{-31} kg\) is the mass of the electron, and \(v\) is its velocity. We know that \(K.E=8.00×10^{-16} J\). Now, rearrange the formula to solve for \(v\): \(v=\sqrt{\frac{2K.E}{m_e}}\) Plug in the values: \(v=\sqrt{\frac{2(8.00×10^{-16} J)}{9.11×10^{-31} kg}}\) \(v= 1.33×10^6 m/s\) Now, compare the speed (v) with 10% of the speed of light, which is 0.1c: \(0.1c =0.1(3×10^8 m/s) = 3×10^7 m/s\) Since \(1.33×10^6 m/s\) is less than 10% of the speed of light (\(3×10^7 m/s\)), the electron is not moving at a relativistic speed.

(c) Electron's total energy and momentum

First, let's compute the relativistic total energy (E) and momentum (p) of the electron, with the relations: \(E=\frac{m_ec^2}{\sqrt{1-\frac{v^2}{c^2}}}\) \(p=\frac{m_ev}{\sqrt{1-\frac{v^2}{c^2}}}\) where \(E\) is the total energy, \(p\) is the momentum, \(m_e = 9.11 × 10^{-31} kg\) is the mass of the electron and \(v = 1.33×10^6 m/s\) is the speed of the electron. Plug in the values: \(E=\frac{9.11 ×10^{-31} kg(3×10^8 m/s)^2}{\sqrt{1-\frac{(1.33×10^6 m/s)^2}{(3×10^8 m/s)^2}}} = 8.19 \times 10^{-14} J\) \(p=\frac{9.11 ×10^{-31} kg(1.33×10^6 m/s)}{\sqrt{1-\frac{(1.33×10^6 m/s)^2}{(3×10^8 m/s)^2}}} =2.71 \times 10^{-24} kg\cdot m/s\) Now, let's compute the nonrelativistic total energy (E') and momentum (p'): \(E'=K.E+m_ec^2\) \(p'=m_ev\) Plug in the values: \(E'=8.00 \times 10^{-16} J +(9.11 ×10^{-31} kg(3×10^8 m/s)^2) = 8.19 \times 10^{-14} J\) \(p'=(9.11 ×10^{-31} kg)(1.33×10^6 m/s)=2.71 \times 10^{-24} kg\cdot m/s\) The relativistic and nonrelativistic values for both total energy and momentum are: Total energy: \(E = E' = 8.19 \times 10^{-14} J\) Momentum: \(p = p' = 2.71 \times 10^{-24} kg\cdot m/s\) Although the electron is not moving at a relativistic speed, the relativistic and nonrelativistic values for both energy and momentum do not differ significantly in this case.