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Chapter 19: Problem 156

If \(M_{O}\) is the mass of an oxygen isotope \({ }_{8} \mathrm{O}^{17}, M_{P}\) and are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is [2007] (A) \(\left(M_{O}-17 M_{N}\right) c^{2}\) (B) \(\left(M_{O}-8 M_{P}\right) c^{2}\) (C) \(\left(M_{O}-8 M_{P}-9 M_{N}\right) c^{2}\) (D) \(M_{O} c^{2}\)

Short Answer

Expert verified
The short answer is: (C) \(\left(M_{O}-8 M_{P}-9 M_{N}\right) c^{2}\)

Step by step solution

01

Identify the composition of the isotope

An oxygen isotope with mass number 17 (\({ }_{8} \mathrm{O}^{17}\)) is composed of 8 protons and 17-8=9 neutrons.
02

Analyze the given options

Let's analyze each option one by one: (A) \(\left(M_{O}-17 M_{N}\right) c^{2}\): This option implies that the nuclear binding energy depends only on the mass difference between the oxygen isotope and 17 neutrons. However, we know the isotope consists of 8 protons and 9 neutrons. Hence this option is incorrect. (B) \(\left(M_{O}-8 M_{P}\right) c^{2}\): This option implies that the nuclear binding energy depends only on the mass difference between the oxygen isotope and 8 protons. However, we know the isotope consists of both protons and neutrons. Hence this option is incorrect. (C) \(\left(M_{O}-8 M_{P}-9 M_{N}\right) c^{2}\): This option states that the nuclear binding energy is the mass difference between the oxygen isotope and the combined mass of its constituent 8 protons and 9 neutrons. This option follows the composition we determined in Step 1 and seems correct. (D) \(M_{O} c^{2}\): This option implies that the nuclear binding energy depends only on the mass of the oxygen isotope. However, the binding energy should be related to the mass difference between the isotope and its protons and neutrons. Hence this option is incorrect.
03

Confirm the correct answer

As per our analysis, option C agrees with the composition of the oxygen isotope and its nuclear binding energy definition. Therefore, the correct answer is: (C) \(\left(M_{O}-8 M_{P}-9 M_{N}\right) c^{2}\)

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Key Concepts

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

Understanding Nuclear Binding Energy for JEE MAIN
Nuclear binding energy is a crucial concept in the field of physics, especially for students preparing for competitive exams like JEE MAIN. In essence, it refers to the energy required to split a nucleus into its constituent protons and neutrons. This energy comes from the mass defect, which is the difference between the mass of the combined nucleus and the sum of the individual masses of the protons and neutrons.

For JEE MAIN aspirants, grasping the calculation of nuclear binding energy is vital. It involves applying the mass-energy equivalence principle, encapsulated by Einstein鈥檚 iconic equation, \(E=mc^2\), where \(E\) is energy, \(m\) is mass, and \(c\) is the speed of light.
The Significance of Oxygen Isotopes
Isotopes are variants of a particular chemical element that have the same number of protons but different numbers of neutrons. Oxygen isotopes, like the one detailed in our problem (\(\_{8}O^{17}\)), have key applications in both basic and applied sciences. Notably, oxygen isotopes help us understand climatological patterns, since they play a part in paleoclimatology studies through ice core sampling.

Students who encounter problems involving oxygen isotopes need to pay attention to isotopic composition, bearing in mind that the atomic mass indicated in the isotope symbol combines both protons and neutrons. It's these subtle differences in mass that influence the calculations related to nuclear binding energies.
Mass-Energy Equivalence in Physics
At the core of many physics problems is the mass-energy equivalence concept introduced by Albert Einstein. This principle asserts that mass can be converted into energy and vice versa, as described by the equation \(E=mc^2\). This revolutionary concept is especially pertinent when discussing nuclear reactions, where extremely small mass changes result in a significant release of energy.

The application of mass-energy equivalence in solving problems, like the one concerning the oxygen isotope鈥檚 nuclear binding energy, entails recognizing how mass loss in the nucleus after binding (mass defect) equates to the binding energy itself. This understanding is paramount for accurately solving related questions, whether in homework exercises or high-stakes exams like JEE MAIN.

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Most popular questions from this chapter

The binding energies of the atoms of elements \(A\) and \(B\) are \(E_{a}\) and \(E_{b}\) respectively. Three atoms of the element \(B\) fuse to give one atom of element \(A\). This fusion process is accompanied by release of energy \(e\). Then \(E_{a}\), \(E_{b}\) and \(e\) are related to each other as (A) \(E_{a}+e=3 E_{b}\) (B) \(E_{a}=3 E_{b}\) (C) \(E_{a}-e=3 E_{b}\) (D) \(E_{a}+3 E_{b}+e=0\)

Pick the correct statements (A) gravitational force between two protons may be greater than the electrostatic force between the protons. (B) electromagnetic force is greater than the gravitational force between two protons. (C) electrostatic force is a fundamental force. (D) nuclear force is attractive or repulsive as per the nature of charges.

In an \(x\)-ray set up accelerating potential difference is set at \(V\) volt and first a target metal having atomic number \(Z_{1}\) is used then target metal having \(Z_{2}\) is used \(\left(\lambda_{\alpha_{1}}\right.\) is wavelength of \(K_{\alpha} x\)-ray from \(Z_{1}\) target and \(\lambda_{\alpha,}\) is wavelength of \(K_{\alpha}, x\)-ray with \(Z_{2}\) target \()\) (A) If \(Z_{1}>Z_{2}\) then \(\left(\lambda_{\alpha_{1}}-\lambda_{\min }\right)>\left(\lambda_{\alpha_{1}}-\lambda_{\min }\right)\) (B) If \(Z_{1}\left(\lambda_{\alpha_{3}}-\lambda_{\min }\right)\) (C) The difference \(\left(\lambda_{\alpha_{1}}-\lambda_{\min }\right)\) and \(\left(\lambda_{\alpha_{12}}-\lambda_{\min }\right)\) will be same for both the target metals. (D) If \(Z_{1} \leq Z_{2}\) then \(\left(\lambda_{\alpha_{1}}-\lambda_{\min }\right) \leq\left(\lambda_{\alpha_{2}}-\lambda_{\min }\right)\)

Uranium ores contain one radium-226 atom for every \(2.8 \times 10^{6}\) Uranium-238 atoms. Calculate the half-life of \({ }_{92} \mathrm{U}^{238}\) given that the half-life of \({ }_{88} \mathrm{Ra}^{226}\) is 1600 years and \({ }_{88} \mathrm{Ra}^{226}\) is a decay product of \({ }_{92} \mathrm{U}^{238}\). (A) \(1.75 \times 10^{3}\) years (B) \(1600 \times \frac{238}{92}\) years (C) \(4.5 \times 10^{9}\) years (D) \(1600 \times \frac{92}{238}\) years

If the maximum kinetic energy of emitted photo electrons from a metal surface of work function \(2.5 \mathrm{eV}\), is \(1.7 \mathrm{eV}\). If wavelength of incident radiation is halved, then stopping potential will be (A) \(2.5 \mathrm{~V}\) (B) \(6.7 \mathrm{~V}\) (C) \(5 \mathrm{~V}\) (D) \(1.1 \mathrm{~V}\)
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