/*! 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 149 A nuclear transformation is deno... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A nuclear transformation is denoted by \(X(n, \alpha)_{3}^{7} L i\). Which of the following is the nucleus of element \(x\) ? [2005] (A) \({ }_{5}^{10} \mathrm{Be}\) (B) \({ }^{12} C_{6}\) (C) \({ }_{4}^{11} B e\) (D) \({ }_{5}^{9} B\)

Short Answer

Expert verified
The nucleus of element X is \(_{5}^{10} B e\).

Step by step solution

01

Understand the nuclear transformation notation

The nuclear transformation is given as \(X(n, \alpha)_{3}^{7} L i\). This can be interpreted as follows: - X is the unknown nucleus of element we need to find. - \(n\) represents a neutron involved in the reaction. - \(\alpha\) is an alpha particle (which consists of 2 protons and 2 neutrons). - \(_{3}^{7} L i\) is the resulting lithium nucleus, with 3 protons and, in total, 7 nucleons (protons and neutrons).
02

Write a balanced nuclear equation

Since the transformation involves an alpha particle, we can write the balanced nuclear equation for this transformation as: \[X + _{0}^{1} n \to _{3}^{7} L i + _{2}^{4} \alpha\]
03

Calculate the number of protons and nucleons in element X

To calculate the number of protons and nucleons (protons and neutrons) in element X, we need to balance the total number of protons and nucleons in both sides of the reaction: For protons: - In element X, let's denote its number of protons by Z. - In the balanced equation, we have Z (in element X) + 0 (in neutron) = 3 (in lithium) + 2 (in alpha particle) For nucleons: - In element X, let's denote its total number of nucleons by A. - In the balanced equation, we have A (in element X) + 1 (in neutron) = 7(in lithium) + 4 (in alpha particle)
04

Solve for the unknowns

Now, let's solve the equations in step 3 to find the values for Z and A: For protons: - \(Z + 0 = 3 + 2\) - \(Z = 5\) For nucleons: - \(A + 1 = 7 + 4\) - \(A = 10\) We find that the nucleus of element X is \(_{5}^{10} B e\).
05

Identify the corresponding option

Now that we have found the nucleus of element X, we need to match it with one of the given options. The correct option is: (A) \(_{5}^{10} \mathrm{Be}\)

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alpha Decay
Alpha decay is a type of nuclear transformation where an unstable nucleus emits an alpha particle to become more stable. Alpha particles consist of two protons and two neutrons, which are the same as a helium nucleus. When an atom undergoes alpha decay, its atomic number decreases by two and its mass number decreases by four.
This means elements become more stable by losing a small, dense part of their nucleus.For example, in our exercise, the unknown nucleus loses an alpha particle that is represented as \(_{2}^{4} \alpha\). This loss affects the balance of both protons and neutrons in the nucleus, which requires compensating changes as seen in nuclear reactions.
Alpha decay is common in heavy elements, where this process helps them reach a more stable state. It is important to understand this concept as part of nuclear reactions in general.
Nuclear Reactions
Nuclear reactions involve changes in an atom's nucleus and can release or absorb extraordinary amounts of energy. These reactions differ from typical chemical reactions, which involve only electrons and are less energetic.
An important aspect of nuclear reactions is the balancing of protons and nucleons, which must be the same on both sides of the equation for nucleons to be conserved.In our example, the equation \(X + _{0}^{1} n \to _{3}^{7} Li + _{2}^{4} \alpha\) demonstrates a complete nuclear reaction. You can see how the neutron \(_{0}^{1} n\) helps balance the nuclear transformation allowing the lithium atom to form. The result is a new nucleus once the reaction balances protons and nucleons.
Balancing is crucial because it ensures the conservation of mass number and atomic number, which is mandatory for accurately predicting the products of such reactions.This deep understanding of nuclear reactions is fundamental to interpreting various phenomena like energy generation in the sun, nuclear power, and artificial transformation in laboratories.
Protons and Nucleons
Protons and nucleons are key terms in understanding nuclear physics. Protons carry a positive charge and define the identity of an element (its atomic number). Nucleons include both protons and neutrons, which comprise the nucleus of an atom.
Neutrons have no charge, contributing to the atomic mass but not changing the chemical element.In our solution, these concepts are crucial for determining the identity of the unknown nucleus \(X\). By setting up balance equations, we solve for the number of protons \(Z\) and nucleons \(A\) in the element using the rules of addition and conservation in nuclear reactions.
Using the balanced nuclear transformation equation, we identified that \(Z = 5\) and \(A = 10\). These results helped us recognize the unknown nucleus as \({}_{5}^{10} \mathrm{Be}\), or Beryllium.
Understanding how protons and nucleons interact and balance during transformations allows us to identify unknown elements and predict reaction outcomes in nuclear chemistry. Thus, mastering these basics is quintessential for delving deeper into nuclear transformations.

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

An electron in hypothetical hydrogen atom is in its \(3^{\text {rd }}\) excited state and makes transition from \(3^{\text {rd }}\) to \(2^{\text {nd }}\) excited, then to \(1^{\text {st }}\) excited state and then to ground state. If the amount of time spent by the electron in any state of quantum number \(n\), is proportional to \(\left(\frac{1}{n-1}\right)\), then the ratio of number of revolutions completed by the electron in \(1^{\text {st }}\) excited state to that in the \(2^{\text {nd }}\) excited state will be (A) 2 (B) \(\frac{27}{8}\) (C) \(\frac{27}{4}\) (D) \(\frac{27}{6}\)

Assertion: Work function of aluminium is \(4.2 \mathrm{eV}\). If two photons of each of energy \(2.5 \mathrm{eV}\) strike on an electron of aluminium, the electron is not emitted. Reason: In photoelectric effect a single photon interacts with a single electron and electron is emitted only if energy of each of incident photon is greater than the work function. (A) \(\mathrm{A}\) (B) B (C) C (D) D

The work function \(W_{A}\) for photoelectric material \(A\) is \(2 \mathrm{eV}\) and \(W_{B}\) for another photoelectric material \(B\) is \(4 \mathrm{eV}\). If photons of energy \(E_{A}\) strike the surface of \(A\), the ejected photoelectrons have a minimum de Broglie wavelength and photons of energy \(E_{B}\) strike the surface \(B\), the ejected photoelectrons also have a minimum de Broglie's wavelength. If \(E_{B}-E_{A}=\) \(0.5 \mathrm{eV}\) and \(V_{A}\) and \(V_{B}\) are the respective stopping potentials, find \(V_{A}-V_{B}\).

Match List-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the choices given below the list \([\mathbf{2 0 1 5}]\) List-I (a) Franck-Hertz Experiment (b) Photo-Electric Experiment (c) Davison-Germer Experiment List-II (i) Particle Nature of Light (ii) Discrete Energy Levels of Atom (iii) Wave nature of Electron (iv) Structure of Atom (A) (a) - (ii), (b) - (iv), (c) - (iii) (B) (a) - (ii), (b) - (i), (c) - (iii) (C) (a) - (iv), (b) - (iii), (c) - (ii) (D) (a) - (i), (b) - (iv), (c) - (iii)

Two radioactive substances \(X\) and \(Y\) initially contain equal number of nuclei. \(X\) has a half-life of 1 hour and \(Y\) has half-life of 2 hours. After two hours the ratio of the activity of \(X\) to the activity of \(Y\) will be (A) \(1: 4\) (B) \(1: 2\) (C) \(1: 1\) (D) \(2: 1\)

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.