What is the momentum of an electron traveling at \(0.980 c ?\)

We are given: - The velocity of the electron (v) is 0.980c - The mass of the electron (m) is approximately 9.11 x 10^(-31) kg. The relativistic momentum formula is: \( p = \dfrac{mv}{\sqrt{1 - \dfrac{v^2} {c^2}}} \)

The given velocity is 0.980c. We must convert it into meters per second (m/s) by multiplying the velocity by the speed of light (c = 3.00 × 10^8 m/s). v = 0.980c = (0.980)(3.00 × 10^8 m/s) ≈ 2.94 × 10^8 m/s

Now, we'll substitute the values of mass (m), velocity (v), and the speed of light (c) into the relativistic momentum formula: \( p = \dfrac{(9.11 \times 10^{-31} \text{ kg})(2.94 \times 10^{8} \text{ m/s})}{\sqrt{1 - \dfrac{(2.94 \times 10^{8} \text{ m/s})^2}{(3.00 \times 10^{8} \text{ m/s})^2}}} \)

Now, we'll first find the squared value in the denominator and then complete the calculations to solve for p: \( p = \dfrac{(9.11 \times 10^{-31} \text{ kg})(2.94 \times 10^{8} \text{ m/s})}{\sqrt{1 - \dfrac{(2.94 \times 10^{8} \text{ m/s})^2}{(3.00 \times 10^{8} \text{ m/s})^2}}} \) \( p = \dfrac{(9.11 \times 10^{-31} \text{ kg})(2.94 \times 10^{8} \text{ m/s})}{\sqrt{1 - \dfrac{(8.64 \times 10^{16} \text{ m^2/s^2)}}{(9.00 \times 10^{16} \text{ m^2/s^2})}}} \) \( p = \dfrac{(9.11 \times 10^{-31} \text{ kg})(2.94 \times 10^{8} \text{ m/s})}{\sqrt{1 - 0.960}} \) \( p = \dfrac{(9.11 \times 10^{-31} \text{ kg})(2.94 \times 10^{8} \text{ m/s})}{\sqrt{0.040}} \) \( p \approx \dfrac{(9.11 \times 10^{-31} \text{ kg})(2.94 \times 10^{8} \text{ m/s})}{0.20} \) \( p \approx 1.34 × 10^{-22} \text{ kg m/s} \) So, the momentum of the electron traveling at 0.980c is approximately 1.34 × 10^(-22) kg m/s.