/*! 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 6 When \({ }_{3} \mathrm{Li}^{7}\)... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 6

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) Neutrons. (B) Alpha particles. (C) Beta particles. (D) Gamma photons.

Short Answer

Expert verified
The emitted particles in this reaction are (D) Gamma photons.

Step by step solution

01

Write down the given nuclear reaction

In this step, we will write down the given nuclear reaction which is: \[_3^7 \textrm{Li} + _1^1 \textrm{p} \rightarrow _4^8 \textrm{Be} + ?\]
02

Check the conservation of atomic numbers (protons) and mass numbers (nucleons)

Let's check if the atomic numbers (protons) and mass numbers (nucleons) are conserved in this reaction. On the left side of the reaction, we have: - Atomic number: 3 (Li) + 1 (p) = 4 - Mass number: 7 (Li) + 1 (p) = 8 On the right side, we have: - Atomic number: 4 (Be) - Mass number: 8 (Be) As we can see, both atomic and mass numbers are conserved in this reaction. This means no particle was emitted in this reaction, and the correct answer is (D) Gamma photons. The complete nuclear reaction is: \[_3^7 \textrm{Li} + _1^1 \textrm{p} \rightarrow _4^8 \textrm{Be} + \gamma\]

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Most popular questions from this chapter

An electron and a proton are separated by a large distance. The electron starts approaching the proton with energy \(2 \mathrm{eV}\). The proton captures the electron and forms a hydrogen atom in first excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength \(4600 \AA\). The maximum \(K E\) of the emitted photoelectron is \((h c=12420 \mathrm{eV} \AA)\) (A) \(2.4 \mathrm{eV}\) (B) \(2.7 \mathrm{eV}\) (C) \(2.9 \mathrm{eV}\) (D) \(5.4 \mathrm{eV}\)

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 \(V_{A}-V_{B}\).

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) \(4.24 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

The energy that should be added to an electron, to reduce its de-Broglie wavelengths from \(10^{-10} \mathrm{~m}\) to \(0.5 \times 10^{-10} \mathrm{~m}\), will be (A) four times the initial energy. (B) thrice the initial energy. (C) equal to the initial energy. (D) twice the initial energy.

An electron and a proton are separated by a large distance. The electron starts approaching the proton with energy \(2 \mathrm{eV}\). The proton captures the electron and forms a hydrogen atom in first excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength \(4600 \AA\). The maximum \(K E\) of the emitted photoelectron is (Take \(h c=12420 \mathrm{eV} \AA\) ) (A) \(2.4 \mathrm{eV}\) (B) \(2.7 \mathrm{eV}\) (C) \(2.9 \mathrm{eV}\) (D) \(5.4 \mathrm{eV}\)
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