/*! 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 129 At a specific instant emission o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 129

At a specific instant emission of radioactive compound is deflected in a magnetic field. The compound can emit: (i) Electrons (ii) Protons (iii) \(\mathrm{He}^{2+}\) (iv) Neutrons

Short Answer

Expert verified
In the presence of a magnetic field, (i) electrons, (ii) protons, and (iii) \(\mathrm{He}^{2+}\) ions will be deflected due to the Lorentz force experienced as a result of their charges. However, (iv) neutrons, being neutral particles with no charge, will not experience any force in the magnetic field and will not be deflected.

Step by step solution

01

Recall the force on a charged particle in a magnetic field

When a charged particle moves through a magnetic field, it experiences a force given by the Lorentz force equation: \( \vec{F} = q(\vec{v} \times \vec{B}) \) Here, \(\vec{F}\) is the force experienced by the particle, \(q\) is the charge of the particle, \(\vec{v}\) is the velocity of the particle, and \(\vec{B}\) is the magnetic field vector.
02

Determine the effect of magnetic field on each type of emitted particle

Now, let's determine the force experienced by each of these particles in the magnetic field when they are emitted. (i) Electrons: Electrons are negatively charged particles with charge \(q = -e\). Therefore, they experience a force in the magnetic field and will be deflected. (ii) Protons: Protons are positively charged particles with charge \(q = +e\). Therefore, they also experience a force in the magnetic field and will be deflected. (iii) \(\mathrm{He}^{2+}\): This is a helium ion with a charge of \(q = +2e\). Thus, it will also experience a force in the magnetic field and will be deflected. (iv) Neutrons: Neutrons are neutral particles with no charge (\(q = 0\)). Hence, they do not experience any force in the magnetic field, and they will not be deflected.
03

Summarize the deflections of the emitted particles

To summarize, we have: (i) Electrons: Deflected; (ii) Protons: Deflected; (iii) \(\mathrm{He}^{2+}\): Deflected; (iv) Neutrons: Not deflected.

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 91影视!

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 particle of mass \(3 m\) at rest decays into two particles of masses \(m\) and \(2 m\) having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles \(\left(\lambda_{1} / \lambda_{2}\right)\) is (A) \(\frac{1}{2}\) (B) \(\frac{1}{4}\) (C) 2 (D) None of these

A radioactive sample at any instant has its disintegration rate 5000 disintegration per minute. After 5 minute, the rate is 1250 disintegrations per minute. Then, the decay constant (per minute) is (A) \(0.4 \ln 2\) (B) \(0.2 \ln 2\) (C) \(0.1 \ln 2\) (D) \(0.8 \ln 2\)

In Bohr's model, the atomic radius of the first orbit is \(r_{0}\), then the radius of the third orbit is (A) \(\frac{r_{0}}{9}\) (B) \(r_{0}\) (C) \(9 r_{0}\) (D) \(3 r_{0}\)

The wavelength involved in the spectrum of deuterium \(\left({ }_{1}^{2} D\right)\) are slightly different from that of hydrogen spectrum because (A) the size of two nuclei are different. (B) the nuclear forces are different in the two cases. (C) the masses of the two nuclei are different. (D) the attraction between the electron and the nuclei is different in the two cases.

When photons of energy \(5 \mathrm{eV}\) strike the surface of a metal \(A\), the ejected photoelectrons have maximum kinetic energy \(K_{A} \mathrm{eV}\) and de Broglie wavelength \(\lambda_{A}\). The maximum kinetic energy of photoelectrons liberated from another metal \(B\) by photons of energy \(5.30 \mathrm{eV}\) is \(K_{B}=\left(K_{A}-1.5\right) . \mathrm{eV}\). If the de Broglie wavelength of these photoelectrons is \(\lambda_{B}=2 \lambda_{A}\), then find \(K_{A}\) and \(K_{B}\).
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Modern Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Fluids

Read Explanation

Scientific Method Physics

Read Explanation

Work Energy and Power

Read Explanation
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.