/*! 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 144 An \(\alpha\)-particle of energy... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 144

An \(\alpha\)-particle of energy \(5 \mathrm{MeV}\) is scattered through \(180^{\circ}\) by a fixed uranium nucleus. The distance of closet approach is of the order of (A) \(10^{-12} \mathrm{~cm}\) (B) \(10^{-10} \mathrm{~cm}\) (C) \(1 \mathrm{~A}\) (D) \(10^{-15} \mathrm{~cm}\)

Short Answer

Expert verified
The distance of closest approach of an $\alpha$-particle with energy $5 \mathrm{MeV}$ scattered through $180^{\circ}$ by a fixed uranium nucleus is of the order of \(10^{-12}\,\mathrm{~cm}\). Hence, the correct option is (A).

Step by step solution

01

Calculate the electrostatic force between alpha particle and uranium nucleus

The force between two charged particles can be expressed as, \(F = k \frac{q_1 q_2}{r^2}\) where k is the electrostatic constant, q1 and q2 are the charges of alpha particle (伪) and uranium nucleus (U), and r is the distance between them. Alpha particle has a charge of +2e, and uranium nucleus (U-238) has a charge of 92e, where e is the elementary charge, e = 1.6 脳 10鈦宦光伖 C. So, the electrostatic force between them is given by: \(F = k \frac{(2e)(92e)}{r^2}\)
02

Calculate the potential energy required for closest approach

The potential energy required for closest approach can be calculated using the relation, \(V = k \frac{q_1 q_2}{r}\) At the closest approach, the kinetic energy of alpha particle will be zero, so the initial kinetic energy will be fully converted into potential energy. Hence, the potential energy at the closest approach can be found by equating the initial energy of the alpha particle to the potential energy. \(E_{initial} = V\), where E_initial is 5 MeV or 5 脳 10鈦� eV. We can now equate the required potential energy with the initial energy of the alpha particle: \(k \frac{(2e)(92e)}{r} = 5 脳 10鈦� eV\)
03

Use the energy conservation principle to find the distance of closest approach

Now, we can use the energy conservation principle to find the closest approach distance (r). Solving for r in the equation obtained in Step 2: \(r = k \frac{(2e)(92e)}{5 脳 10鈦� eV}\), where \(k = \frac{1}{4\pi\epsilon_0} = 8.988\times 10^9\,Nm^2/C^2\). \(r = \frac{(2)(92)(8.988\times 10^9)(1.6\times 10^{-19})^2}{5\times 10^6(1.6\times 10^{-19})}\) Last step is to calculate r which results in: \(r \approx 10^{-14}\,\mathrm{m}\) Converting to cm: \(r \approx 10^{-12}\,\mathrm{~cm}\) So, the distance of closest approach is of the order of \(10^{-12}\,\mathrm{~cm}\). Hence, our answer is (A).

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 nucleus with \(Z=92\) emits the following in a sequence: \(\alpha, \beta^{-}, \beta^{-} \alpha, \alpha, \alpha, \alpha, \alpha, \beta^{-}, \beta^{-}, \alpha, \beta^{+}, \beta^{+}, \alpha\) Then \(z\) of the resulting nucleus is (A) 76 (B) 78 (C) 82 (D) 74

Which of the following is a correct statement? (A) Beta rays are same as cathode rays. (B) Gamma rays are high energy neutrons. (C) Alpha particles are double ionized helium atoms. (D) Protons and neutrons have exactly the same mass.

As an electron makes a transition from an excited state to the ground state of hydrogen - like atom/ion \([2015]\) (A) \(K E\), potential energy and total energy decrease. (B) \(K E\) decreases, potential energy increases but total energy remains same. (C) \(K E\) and total energy decrease but potential energy increases. (D) Its kinetic energy increases but potential energy and total energy decrease.

The half-life of a radioactive substance is \(20 \mathrm{~min}\). The approximate time interval \(\left(t_{2}-t_{1}\right)\) between the time \(t_{2}\) when \(\frac{2}{3}\) of it has decayed and time \(t_{1}\) when \(\frac{1}{3}\) of it had decayed is (A) \(7 \mathrm{~min}\) (B) \(14 \mathrm{~min}\) (C) \(20 \mathrm{~min}\) (D) \(28 \mathrm{~min}\)

The binding energy per nucleon of deuteron \(\left({ }_{1}^{2} \mathrm{H}\right)\) and helium nucleus \(\left({ }_{2}^{4} \mathrm{He}\right)\) is \(1.1 \mathrm{MeV}\) and \(7 \mathrm{MeV}\) respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is (A) \(23.6 \mathrm{MeV}\) (B) \(26.9 \mathrm{MeV}\) (C) \(13.9 \mathrm{MeV}\) (D) \(19.2 \mathrm{MeV}\)
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Fields in Physics

Read Explanation

Scientific Method Physics

Read Explanation

Translational Dynamics

Read Explanation

Dynamics

Read Explanation

Physical Quantities And Units

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.