/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 In a medical \(X\)-ray tube, ele... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a medical \(X\)-ray tube, electrons are accelerated to a velocity of \(10^{8} \mathrm{~m} / \mathrm{s}\) and then slammed into a tungsten target. As they stop, the electrons' rapid acceleration produces X rays. Given that it takes an electron on the order of \(1 \mathrm{~ns}\) to stop, estimate the distance it moves while stopping.

Short Answer

Expert verified
Thus, the estimated distance that an electron moves while stopping is \(5 \times 10^{-2} \mathrm{m}\) or 5 cm.

Step by step solution

01

Understand the concept of deceleration

Deceleration is the opposite of acceleration, it is a decrease in velocity. In this case, the electron is decelerating rapidly, going from its initial velocity to a final velocity of zero in a very short time.
02

Convert nanoseconds to seconds

Time is given in nanoseconds. To maintain unit consistency, convert it into seconds as the velocity is given in meters per second. Note that \(1 \mathrm{~ns} = 1 \times 10^{-9} \mathrm{~s}\). So, the time is \(1 \times 10^{-9}\) seconds.
03

Calculate the distance travelled

We know that distance traveled is equal to average velocity multiplied by time, where average velocity is the sum of initial and final velocity divided by 2. In this case, the initial velocity is \(10^{8} \mathrm{~m} / \mathrm{s}\) and the final velocity is 0. To find the total distance traveled, the formula \(d = \frac{{v_i + v_f}}{2} \cdot t\) is used. Substituting the values \(v_i = 10^8 \mathrm{m/s}\), \(v_f = 0\), and \(t = 1 \times 10^{-9} \mathrm{s}\) into the formula we get \(d = \frac{{10^8 + 0}}{2} \times 1 \times 10^{-9}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Electron Deceleration
The phenomenon of electron deceleration is integral to grasping how an X-ray tube operates. Imagine this as a car braking from a high speed to a full stop; similarly, electrons in an X-ray tube slow down abruptly from a high velocity. This deceleration is key to producing X-ray photons.

In the medical X-ray tube, electrons are shot towards a tungsten target at incredibly high speed—on the level of hundreds of millions of meters per second. When these electrons hit the tungsten material, they don't just stop instantaneously; there's a tiny, yet measurable distance over which they decelerate to a halt.

Why tungsten, you might ask? Tungsten is chosen because it's a metal with a very high melting point and the ability to withstand intense heat without damage, making it an ideal material for this high-energy collision process that generates useful X-ray photons for medical imaging.
Estimating the Distance of Deceleration
Calculating the distance an electron travels as it slows down within an X-ray tube is a classic physics problem, akin to figuring out how far a car travels as it comes to a stop. To determine this, we consider the electron's initial speed (a whopping 100 million meters per second) and the brief time it takes to decelerate to rest.

To visualize this scenario, picture the electron as a sprinter on a track who suddenly needs to stop. The time it takes for the stop is incredibly short, just a nanosecond, but during this nanosecond, the electron still covers some ground.

The average velocity is a crucial piece in this puzzle—it's the average speed of the electron during its deceleration phase. Since we know the electron decelerates to rest, we calculate the average of the initial velocity and the final velocity, which is zero. To find the stopping distance, we simply multiply this average velocity by the time interval of deceleration—giving us our estimated travel distance while stopping.
From Nanoseconds to Seconds: Handling Time Units
In physics, harmonizing units is fundamental to achieve correct calculations. The time it takes for the electron to stop is given as one nanosecond—that's one billionth of a second! The challenge here is to convert this minuscule fraction into seconds for consistency with the velocity given in meters per second.

Understanding the prefix 'nano' is crucial. 'Nano-' denotes a factor of one-billionth, which in the language of exponents converts to a multiplier of 10 to the power of negative nine. Therefore, when we see one nanosecond, we convert this into \( 1 \times 10^{-9} \) seconds.

In the context of our X-ray electron problem, converting time from nanoseconds to seconds is what allows us to utilize the standard physics formulae that require time in seconds, thereby enabling us to compute the distance traveled as the electron decelerates to a stop within the X-ray tube.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A Frisbee is lodged in a tree \(6.4 \mathrm{~m}\) above the ground. A rock thrown from below must be going at least \(3 \mathrm{~m} / \mathrm{s}\) to dislodge the Frisbee. How fast must such a rock be thrown upward if it leaves the thrower's hand \(1.3 \mathrm{~m}\) above the ground?

A motorist suddenly notices a stalled car and slams on the brakes, slowing at \(6.3 \mathrm{~m} / \mathrm{s}^{2}\). Unfortunately, this isn't enough, and a collision ensues. From the damage sustained, police estimate that the car was going \(18 \mathrm{~km} / \mathrm{h}\) at the time of the collision. They also measure skid marks \(34 \mathrm{~m}\) long. (a) How fast was the motorist going when the brakes were first applied? (b) How much time elapsed from the initial braking to the collision?

An airplane's takeoff speed is \(320 \mathrm{~km} / \mathrm{h}\). If its average acceleration is \(2.9 \mathrm{~m} / \mathrm{s}^{2}\), how much time is it accelerating down the runway before it lifts off?

You're speeding at \(85 \mathrm{~km} / \mathrm{h}\) when you notice that you're only \(10 \mathrm{~m}\) behind the car in front of you, which is moving at the legal speed limit of \(60 \mathrm{~km} / \mathrm{h}\). You slam on your brakes, and your car slows at the rate of \(4.2 \mathrm{~m} / \mathrm{s}^{2}\). Assuming the other car continues at constant speed, will you collide? If so, at what relative speed? If not, what will be the distance between the cars at their closest approach?

You're at mission control for a rocket launch, deciding whether to let the launch proceed. A band of clouds \(5.1 \mathrm{~km}\) thick extends upward from \(1.1 \mathrm{~km}\) altitude. The rocket will accelerate at \(4.3 \mathrm{~m} / \mathrm{s}^{2}\), and it isn't allowed to be out of sight for more than \(30 \mathrm{~s}\). Should vou allow the launch?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.