/*! 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 19 If the binding energy per nucleo... [FREE SOLUTION] | 91Ó°ÊÓ

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If the binding energy per nucleon in \({ }_{3}^{7} \mathrm{Li}\) and \({ }_{2}^{4} \mathrm{He}\) nuclei are \(5.60 \mathrm{MeV}\) and \(7.06 \mathrm{MeV}\) respectively, then in the reaction \({ }_{1}^{1} \mathrm{H}+{ }_{3}^{7} \mathrm{Li} \rightarrow 2{ }_{2}^{4} \mathrm{He}\) energy of proton must be (A) \(39.2 \mathrm{MeV}\) (B) \(28.24 \mathrm{MeV}\) (C) \(17.28 \mathrm{MeV}\) (D) \(1.46 \mathrm{MeV}\)

Short Answer

Expert verified
The energy of the proton must be \(17.28 \mathrm{MeV}\)

Step by step solution

01

- Calculate the initial binding energy

Find the initial binding energy by multiplying the binding energies per nucleon for lithium-7 and proton by their respective number of nucleons. As proton's binding energy is zero, only lithium-7's total binding energy needs to be calculated. This gives us a total of \(5.60 \mathrm{MeV} * 7 = 39.2 \mathrm{MeV}\) of initial binding energy.
02

- Calculate the final binding energy

Calculate the final binding energy by multiplying the binding energy per nucleon for helium-4 by the total number of nucleons (as two helium-4 nuclei are formed in the reaction). This gives us a total of \(7.06 \mathrm{MeV} * 8 = 56.48 \mathrm{MeV}\) of final binding energy.
03

- Calculate the energy of the proton

Finally, find the energy of the proton. According to the law of conservation of energy, the initial energy (which is the energy of the proton plus the binding energy of lithium-7) should be equal to the final energy (which is the binding energy of two helium-4 nuclei). The energy of the proton can, therefore, be calculated as \(56.48 - 39.2 = 17.28 \mathrm{MeV}\).

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Key Concepts

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

Binding Energy Per Nucleon
In nuclear physics, the binding energy per nucleon is a critical concept that helps explain the stability of a nucleus. It represents the average energy needed to remove a nucleon (either a proton or a neutron) from the nucleus. This energy is crucial for understanding nuclear reactions, such as the one involving lithium-7.

For instance, in the case of lithium-7 \(({}_{3}^{7} ext{Li})\), the binding energy per nucleon is given as \(5.60 \ ext{MeV}\). This means each of the seven nucleons is bound with an average energy of \ 5.60 \ ext{MeV}. The total binding energy of a nucleus is found by multiplying the binding energy per nucleon by the number of nucleons within that nucleus.

Key points to remember:
  • Binding energy per nucleon indicates nuclear stability.
  • Higher binding energy usually means a more stable nucleus.
  • It's a critical factor in determining the energy outcome in nuclear reactions.
Conservation of Energy
The conservation of energy is a fundamental principle that states energy cannot be created or destroyed, only transformed from one form to another. In nuclear reactions, this principle is vital as it ensures that the total energy before and after a reaction remains constant.

In the given exercise, the energy of the proton is linked to the change in binding energy before and after the reaction. Initially, we only consider the binding energy of lithium-7 since a proton's binding energy is zero. The final energy involves the binding energies of two helium-4 nuclei which have a significantly higher combined binding energy.

This change can be explained as:
  • Initial energy consists of the proton energy plus lithium-7 binding energy.
  • Final energy consists only of the helium nuclei binding energies.
  • The difference in these energies is expressed in the kinetic energy of the proton.
Lithium-7
Lithium-7 is an isotope of lithium, composed of three protons and four neutrons, making a total of seven nucleons. It's relatively stable and plays a significant role in nuclear reactions due to its specific binding energy characteristics.

In nuclear reactions, such as the one in the exercise, lithium-7 interacts with other particles like protons to transform into other elements – in this case, helium. This transformation is governed by changes in nucleon arrangement and binding energies.

Important points regarding lithium-7:
  • It serves as a target for nuclear reactions involving light particles.
  • Its interactions often result in the production of more stable nuclei like helium-4.
  • The energy relations in reactions involving lithium-7 are crucial for applications, like energy generation and astrophysics.

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Most popular questions from this chapter

The speed of an electron having a wavelength of the order of \(1 \AA\) will be (A) \(7.25 \times 10^{6} \mathrm{~m} / \mathrm{s}\) (B) \(6.26 \times 10^{6} \mathrm{~m} / \mathrm{s}\) (C) \(5.25 \times 10^{6} \mathrm{~m} / \mathrm{s}\) (D) \(4.24 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its kinetic energy. Consequently the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The minimum kinetic energy of colliding electron is (A) \(10.2 \mathrm{eV}\) (B) \(1.9 \mathrm{eV}\) (C) \(12.09 \mathrm{eV}\) (D) \(13.6 \mathrm{eV}\)

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.

The mean lives of a radioactive sample are 30 years and 60 years for \(\alpha\)-emission and \(\beta\)-emission respectively. If the sample decays both by \(\alpha\)-emission and \(\beta\)-emission simultaneously, the time after which, only one-fourth of the sample remain is (A) 10 years (B) 20 years (C) 40 years (D) 45 years

The bodies \(A\) and \(B\) have thermal emissivity's of \(0.01\) and \(0.81\) respectively. The outer surface areas of the two bodies are equal. The two bodies emit total radiant power at the same rate. The wavelength \(\lambda_{B}\) corresponding to maximum spectral radiance in the radiation from \(B\) is shifted from the wavelength corresponding to maximum spectral radiance in the radiation from \(A\), by \(1.00 \mu \mathrm{m}\). If the temperature of \(A\) is \(5802 \mathrm{~K}\). (A) the temperature of \(B\) is \(1934 \mathrm{~K}\). (B) \(\lambda_{B}=1.5 \mu \mathrm{m}\). (C) the temperature of \(B\) is \(1160 \mathrm{~K}\). (D) the temperature of \(B\) is \(2901 \mathrm{~K}\).

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