Calculating the Speed of an Electron with Kinetic Energy of 1.02 MeV

Understanding the speed of an electron, especially when it has a significant kinetic energy, can be fundamental in both theoretical and applied physics. In this article, we will explore how to determine the speed of an electron with a kinetic energy of 1.02 MeV using principles of relativistic mechanics. We will break down the process step-by-step with a detailed explanation and calculation.

Relativistic Energy-Momentum Relation

For a relativistic particle, the energy-momentum relation is crucial. The equation that describes this is:

E2 (pc2 (mc2)2)

In this equation:

E is the total energy of the electron. p is the momentum of the electron. m is the rest mass of the electron. c is the speed of light.

We can also relate the kinetic energy KE to the total energy E and the rest mass energy mc2 using the following equation:

KE E - mc2

Step-by-Step Calculation

Let's begin the calculation with the given kinetic energy of 1.02 MeV and the rest mass energy of an electron, which is 0.511 MeV.

1. Calculate the Total Energy

First, we need to calculate the total energy E of the electron:

E KE mc2

Given:

KE 1.02 MeV mc2 0.511 MeV

So, the total energy E is:

E 1.02 MeV 0.511 MeV 1.531 MeV

2. Calculate the Momentum

Next, we use the relativistic energy-momentum relation to find the momentum p:

pc2 E2 - (mc2)2

E2 (1.531 MeV)2 2.343961 MeV2 (mc2)2 (0.511 MeV)2 0.261121 MeV2

Now, we can calculate the momentum:

pc2 2.343961 MeV2 - 0.261121 MeV2 2.08284 MeV2 p sqrt(2.08284 MeV2) 1.42 MeV/c

3. Calculate the Lorentz Factor and Speed

To find the speed v, we first need to calculate the Lorentz factor γ:

γ E / mc2 γ 1.531 MeV / 0.511 MeV 3

Finally, we can find the velocity using the relativistic momentum formula:

ρ γmv

Solving for v gives:

v p / (γm) ≈ (1.42 MeV/c) / 3(0.511 MeV/c2) ≈ 0.94c

Thus, the speed of the electron is approximately 0.94 times the speed of light.

Conversion and Alternative Calculation

For a more direct calculation, we can use the relationship between kinetic energy and speed:

KE 0.5mv2}

Given that the electron is moving at relativistic speeds, we need to use the relativistic kinetic energy formula:

KE γmc2}

Where:

γ 1 / sqrt(1 - (v2 / c2)) m ≈ 9.11 × 10-31} kg c ≈ 3.00 × 108} m/s

1. Convert Kinetic Energy to Joules

1 MeV 1.602 × 10-13 J 1.02 MeV 1.02 × 1.602 × 10-13 J ≈ 1.634 × 10-13} J

2. Calculate Total Energy E

Total energy: E KE mc2 mc2 (9.11 × 10-31} × (3.00 × 108}2) 8.187 × 10-14} J

E 1.634 × 10-13} J 8.187 × 10-14} J ≈ 2.452 × 10-13} J

3. Calculate Lorentz Factor γ

γ E / mc2 γ 2.452 × 10-13} J / 8.187 × 10-14} J ≈ 2.997

4. Find Speed v

Using γ 1 / sqrt(1 - (v2 / c2)) 1 - (v2 / c2) 1 / (γ2) ≈ 1 / (2.9972) ≈ 0.1113 (v2 / c2) 1 - 0.1113 ≈ 0.8887 v2 ≈ c2 × 0.8887 v ≈ c × sqrt(0.8887) ≈ 3.00 × 108} m/s × 0.9428 ≈ 2.83 × 108} m/s

Therefore, the speed of the electron is approximately 2.83 × 108} m/s

Conclusion

Through these calculations, we have determined that the speed of an electron with a kinetic energy of 1.02 MeV, a significant amount of energy, is remarkably close to the speed of light. This demonstrates the profound impact of relativistic dynamics on particles with high kinetic energies.