/*! 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} Q26 PE Using values from Table 7.1, how... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Using values from Table 7.1, how many DNA molecules could be broken by the energy carried by a single electron in the beam of an old-fashioned TV tube? (These electrons were not dangerous in themselves, but they did create dangerous x rays. Later model tube TVs had shielding that absorbed x rays before they escaped and exposed viewers.)

Short Answer

Expert verified

A single electron in the beam of old-fashioned TV tube can break 40000molecules of DNA.

Step by step solution

01

Conservation of Energy

Conservation of energy:The total energy of an isolated system is always conserved. In other words, the energy neither be created nor be destroyed; it can be only transformed from one form to another.

02

Number of molecules of DNA can break by a single electron

When a single electron in the beam of an old-fashioned TV tube strikes on the molecules of DNA, the total energy of the single electron will break many molecules of DNA. Mathematically,

Eelectron=nEDNA

Here,Eelectronis the energy of single electron in the beam of old-fashioned TV tubeEelectron=4.0×10-15J, n is number of molecules of DNA broken, andEDNAis the energy required to break one DNA moleculeEDNA=10-19J.

The expression for the number of molecules of DNA broken is,

n=EelectronEDNA

Putting all known values,

n=4.0×10-15J10-19J=40000

Therefore, a single electron in the beam of anold-fashioned TV tube can break40000 molecules of DNA.

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!

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

Suppose a car travels\(108{\rm{ km}}\)at a speed of\(30.0\,{\rm{m}}/{\rm{s}}\), and uses\(2.0{\rm{ gal}}\)of gasoline. Only\(30\% \)30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (See Table 7.1 for the energy content of gasoline.)

(a) What is the magnitude of the force exerted to keep the car moving at constant speed?

(b) If the required force is directly proportional to speed, how many gallons will be used to drive\(108{\rm{ km}}\) at a speed of \(28.0{\rm{ m}}/{\rm{s}}\)?

(a) Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a non-panic stop).

(b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a).

A car’s bumper is designed to withstand a \(4.0 - {\rm{km}}/{\rm{h}}\) \(\left( {1.1 - {\rm{m}}/{\rm{s}}} \right)\) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses \(0.200{\rm{ m}}\) while bringing a \(900 - {\rm{kg}}\) car to rest from an initial speed of \(1.1{\rm{ m}}/{\rm{s}}\).

Shivering is an involuntary response to lowered body temperature. What is the efficiency of the body when shivering, and is this a desirable value?

List the energy conversions that occur when riding a bicycle.

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.