To determine the electric field strength required to accelerate electrons in an X-ray tube to one-tenth the speed of light, we need to break the problem into steps and use the work-energy principle. Here's the process:

1. Determine the final kinetic energy of the electron.

The kinetic energy $K$ of an electron moving with velocity $v$ is given by the formula:

$K = \frac{1}{2} m v^2$

where:

• $m$ is the mass of the electron ($9.11 \times 10^{-31} \, \text{kg}$),
• $v$ is the final velocity of the electron, which is one-tenth of the speed of light ($v = 0.1 c$).

Since the speed of light $c = 3.00 \times 10^8 \, \text{m/s}$, we have:

$v = 0.1 \times 3.00 \times 10^8 \, \text{m/s} = 3.00 \times 10^7 \, \text{m/s}$

Now, substitute $v$ into the kinetic energy formula:

$K = \frac{1}{2} \times 9.11 \times 10^{-31} \, \text{kg} \times (3.00 \times 10^7 \, \text{m/s})^2$

$K = \frac{1}{2} \times 9.11 \times 10^{-31} \times 9.00 \times 10^{14}$

$K = 4.1 \times 10^{-16} \, \text{J}$

2. Relate the kinetic energy to the work done by the electric field.

The work $W$ done by an electric field $E$ on an electron is given by:

$W = e E d$

where:

• $e$ is the elementary charge of the electron ($1.602 \times 10^{-19} \, \text{C}$),
• $E$ is the electric field strength,
• $d$ is the distance over which the electron is accelerated ($d = 4.5 \, \text{cm} = 0.045 \, \text{m}$).

Since the work done by the electric field equals the kinetic energy, we have:

$e E d = K$

Substitute the values for $e$, $d$, and $K$:

$(1.602 \times 10^{-19} \, \text{C}) \times E \times (0.045 \, \text{m}) = 4.1 \times 10^{-16} \, \text{J}$

3. Solve for the electric field $E$.

Rearrange the equation to solve for $E$:

$E = \frac{K}{e d}$

Substitute the values:

$E = \frac{4.1 \times 10^{-16} \, \text{J}}{(1.602 \times 10^{-19} \, \text{C}) \times (0.045 \, \text{m})}$

$E = \frac{4.1 \times 10^{-16}}{7.209 \times 10^{-21}}$

$E \approx 5.69 \times 10^4 \, \text{N/C}$

Final Answer:

The electric field needed to accelerate the electrons to one-tenth the speed of light in a distance of 4.5 cm is approximately:

$E \approx 5.69 \times 10^4 \, \text{N/C}$

---

Would you like further explanation of any step, or have any questions about the concept? Here's a list of related topics that might interest you:

1. How does the acceleration of an electron depend on the electric field?
2. What is the relationship between energy and speed for relativistic particles?
3. What are the factors that influence the efficiency of an X-ray tube?
4. How is the work-energy theorem applied in particle accelerators?
5. What are the effects of a stronger electric field on electron acceleration?

Tip: When dealing with high-speed particles (like electrons), remember that relativistic effects may become important, but for speeds much lower than the speed of light, the classical formulas still hold.