/*! 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 151 When \({ }_{3} \mathrm{Li}^{7}\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When \({ }_{3} \mathrm{Li}^{7}\) nuclei are bombarded by protons, and the resultant nuclei are \({ }_{4} \mathrm{Be}^{8}\), the emitted particles will be (A) Alpha particles (B) Beta particles (C) Gamma photons (D) Neutrons

Short Answer

Expert verified
None of the given options (A to D) is the correct answer, as the reaction involves an increase in proton count due to bombardment without emitting any of the listed particles.

Step by step solution

01

Understanding the pre-reaction configuration

The nucleus of \( {}_{3} \mathrm{Li}^{7} \) has 3 protons and 4 neutrons (since the symbol represents the isotopes with 3 protons and 7 nucleons (protons+neutrons)). Additionally, we are informed that these nuclei are bombarded by protons.
02

Understanding the post-reaction configuration

The nucleus of \( {}_{4} \mathrm{Be}^{8} \) has 4 protons and 4 neutrons (remember that Be has 4 protons and 8 nucleons).
03

Observing the change in configuration

Comparing the pre and post configurations, we notice that the number of protons has increased by 1 (3 to 4). The number of neutrons remained the same (4 to 4). This means the extra proton from the bombardment has been added to the lithium nucleus, and as no neutron was lost, neutrons can be ruled out as the emitted particle.
04

Understanding the nature of different particles/phenomena

Alpha particles consist of 2 protons and 2 neutrons. Beta particles can be an electron or a positron that is emitted when a neutron in the nucleus is transformed into a proton or vice versa. Gamma photons are released as a form of electromagnetic energy when a nucleus is excited. In this reaction, no such transformations or excitations are taking place. Therefore, alpha particles, beta particles, and gamma photons can also be ruled out.
05

Identifying the emitted particle

From steps 3 and 4, we have effectively ruled out alpha particles, beta particles, gamma photons, and neutrons as potential emitted particles. However, the exercise does not give an option that fits this result, as an increase in proton count due to bombardment does not typically involve the emission of a particle. This seems to be a trick question, and none of the given options (A to D) is the correct answer.

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

Column-I represent the physical parameters being changed in the experiment of photo electric effect and in Column-II is its effect. Column-I (A) Intensity (B) Frequency (C) Potential difference between anode and cathode (D) Metal Column-II (1) photo electric current (2) stopping potential (3) work function (4) maximum kinetic energy

In hydrogen like atom electron makes transition from an energy level with quantum number \(n\) to another with quantum number \((n-1) .\) If \(n \gg>1\), the frequency of radiation emitted is proportional to: (A) \(\frac{1}{n^{2}}\) (B) \(\frac{1}{n^{3 / 2}}\) (C) \(\frac{1}{n^{3}}\) (D) \(\frac{1}{n}\)

Sodium and copper have work done functions \(2.3 \mathrm{eV}\) and \(4.5 \mathrm{eV}\) respectively. Ten the ratio of the wavelengths is nearest to (A) \(1: 2\) (B) \(4: 1\) (C) \(2: 1\) (D) \(1: 4\)

The intensity of gamma radiation from a given source is \(I .\) On passing through \(36 \mathrm{~mm}\) of lead, it is reduced to \(\frac{I}{8} .\) The thickness of lead which will reduce the intensity to \(\frac{I}{2}\) will be (A) \(9 \mathrm{~mm}\) (B) \(6 \mathrm{~mm}\) (C) \(12 \mathrm{~mm}\) (D) \(18 \mathrm{~mm}\)

The de-Broglie wavelength of a particle moving with a velocity \(2.25 \times 10^{8} \mathrm{~m} / \mathrm{s}\) is equal to the wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is (velocity of light is \(\left.3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)\) (A) \(\frac{1}{8}\) (B) \(\frac{3}{8}\) (C) \(\frac{5}{8}\) (D) \(\frac{7}{8}\)

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.