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Chapter 26: Problem 34

An electron has momentum of magnitude $2.4 \times 10^{-22} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .$ What is the electron's speed?

Short Answer

Expert verified
Answer: The electron's speed is approximately \(2.63 \times 10^8 \mathrm{m} / \mathrm{s}\).

Step by step solution

01

Write down the given value and constants.

Here, we have the momentum value (\(p = 2.4 \times 10^{-22} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\)) and the mass of an electron (m_e = \(9.11 \times 10^{-31} \mathrm{kg}\)).
02

Write down the momentum formula.

The formula for momentum is p = mv, where p is momentum, m is mass, and v is velocity.
03

Rearrange the formula to find the velocity.

We can rearrange the momentum formula to isolate the velocity (v). Divide both sides by mass (m): v = p / m
04

Substitute the given values and constants into the formula.

Now, substitute the given values and constants into the formula: v = (2.4 \times 10^{-22} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}) / (9.11 \times 10^{-31} \mathrm{kg})
05

Calculate the velocity.

Perform the calculation: v = (2.4 \times 10^{-22}) / (9.11 \times 10^{-31}) v = 2.63 \times 10^8 \mathrm{m} / \mathrm{s}
06

Final answer.

The electron's speed is approximately \(2.63 \times 10^8 \mathrm{m} / \mathrm{s}\).

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Most popular questions from this chapter

Event A happens at the spacetime coordinates $(x, y, z, t)=(2 \mathrm{m}, 3 \mathrm{m}, 0,0.1 \mathrm{s})$ and event B happens at the spacetime coordinates \((x, y, z, t)=\left(0.4 \times 10^{8} \mathrm{m}\right.\) $3 \mathrm{m}, 0,0.2 \mathrm{s}) .$ (a) Is it possible that event A caused event B? (b) If event B occurred at $\left(0.2 \times 10^{8} \mathrm{m}, 3 \mathrm{m}, 0,0.2 \mathrm{s}\right)$ instead, would it then be possible that event A caused event B? [Hint: How fast would a signal need to travel to get from event \(\mathrm{A}\) to the location of \(\mathrm{B}\) before event \(\mathrm{B}\) occurred?]
Show that each of these statements implies that \(v<c\) which means that \(v\) can be considered a nonrelativistic speed: (a) $\gamma-1<<1 \text { [Eq. }(26-14)] ;\( (b) \)K<<m c^{2}\( [Eq. \)(26-15)] ;(\text { c }) p<m c[\text { Eq. }(26-16)] ;\( (d) \)K \approx p^{2} /(2 m)$.
Suppose that as you travel away from Earth in a spaceship, you observe another ship pass you heading in the same direction and measure its speed to be $0.50 c .$ As you look back at Earth, you measure Earth's speed relative to you to be \(0.90 c .\) What is the speed of the ship that passed you according to Earth observers?
A charged particle is observed to have a total energy of \(0.638 \mathrm{MeV}\) when it is moving at \(0.600 c .\) If this particle enters a linear accelerator and its speed is increased to \(0.980 c,\) what is the new value of the particle's total energy?
Two spaceships are moving directly toward one another with a relative velocity of \(0.90 c .\) If an astronaut measures the length of his own spaceship to be \(30.0 \mathrm{m},\) how long is the spaceship as measured by an astronaut in the other ship?
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