/*! 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} Q2 PE A particle of ionizing radiation... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A particle of ionizing radiation creates \({\rm{4000}}\) ion pairs in the gas inside a Geiger tube as it passes through. What minimum energy was deposited, if \(30.0\,{\rm{eV}}\) is required to create each ion pair?

Short Answer

Expert verified

Maximum energy deposited is obtained as: \(0.12\,{\rm{MeV}}\).

Step by step solution

01

Define Radioactivity

The spontaneous emission of radiation in the form of particles or high-energy photons as a result of a nuclear process is known as radioactivity.

02

Evaluating the minimum energy that was deposited

Number of ions is created as:

\(n = 4 \times {10^3}\,{\rm{ion}}\)

The energy of the radiation in the Geiger tube is given as:

\({E_{rad}} = 30.0\,{\rm{eV}}\)

The number of ion pairs, is obtained using the equation:

\({\rm{Number of ion pairs = }}\frac{{{\rm{The energy of the radiation in the Geiger tube}}}}{{{\rm{The energy for ionize molecule}}}}\)

Rearranging the values and we obtain:

\(\begin{align}E &= n \times {E_{ion}}\\ &= 4 \times {10^3}\,ion \times 30\,eV\\ &= 1.2 \times {10^5}\,eV\\ &= 0.12\,MeV\end{align}\)

Therefore, the minimum energy deposited was: \(0.12\,{\rm{MeV}}\).

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

(a) Calculate the radius of\(^{{\rm{58}}}{\rm{Ni}}\), one of the most tightly bound stable nuclei. (b) What is the ratio of the radius of\(^{{\rm{58}}}{\rm{Ni}}\)to that of\(^{{\rm{258}}}{\rm{Ha}}\), one of the largest nuclei ever made? Note that the radius of the largest nucleus is still much smaller than the size of an atom.

The detail observable using a probe is limited by its wavelength. Calculate the energy of a\({\rm{\gamma }}\)-ray photon that has a wavelength of\(1 \times {10^{ - 16}}\,{\rm{m}}\), small enough to detect details about one-tenth the size of a nucleon. Note that a photon having this energy is difficult to produce and interacts poorly with the nucleus, limiting the practicability of this probe.

Ionizing radiation interacts with matter by scattering from electrons and nuclei in the substance. Based on the law of conservation of momentum and energy, explain why electrons tend to absorb more energy than nuclei in these interactions.

Neutrinos are experimentally determined to have an extremely small mass. Huge numbers of neutrinos are created in a supernova at the same time as massive amounts of light are first produced. When the 1987A supernova occurred in the Large Magellanic Cloud, visible primarily in the Southern Hemisphere and some 100,000 light-years away from Earth, neutrinos from the explosion were observed at about the same time as the light from the blast. How could the relative arrival times of neutrinos and light be used to place limits on the mass of neutrinos?

Unreasonable Results

The manufacturer of a smoke alarm decides that the smallest current of \({\rm{\alpha }}\) radiation he can detect is \(1.00\,\mu A\).

  1. Find the activity in curies of an \({\rm{\alpha }}\) emitter that produces a \(1.00\,\mu A\)current of \({\rm{\alpha }}\) particles.
  2. What is unreasonable about this result?
  3. What assumption is responsible?
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.