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Chapter 18: Problem 2502

In gamma ray emission form a nucleus (A) there is no change in the proton-number and neutron number (B) Both the number are changes (C) only Proton number change (D) only neutron number change

Short Answer

Expert verified
\( \Rightarrow \) The correct answer is (A) There is no change in the proton-number and neutron number during gamma-ray emission from a nucleus.

Step by step solution

01

Understand Gamma-Ray Emission

Gamma-ray emission occurs when a nucleus in an excited state (i.e., with excess energy) transitions to a lower energy state by emitting a high-energy photon known as a gamma-ray. This process does not involve the change of any particles within the nucleus.
02

Analyze the Options

Now, let's analyze each option and determine which one is correct based on our understanding of gamma-ray emissions. (A) No change in the proton-number and neutron number. This statement aligns with our understanding that gamma-ray emission does not involve any changes in the composition of the nucleus. (B) Both proton and neutron numbers change. This statement contradicts our understanding of gamma-ray emissions as they do not involve changes in the composition of the nucleus. (C) Only Proton number change. This statement is also inconsistent with our understanding of gamma-ray emission, as it does not involve any changes in the composition of the nucleus. (D) Only neutron number change. This statement, like the others, contradicts our understanding of gamma-ray emission, as it does not involve any changes in the composition of the nucleus.
03

Choose the Correct Option

Based on our analysis and understanding of gamma-ray emissions, we can conclude that the correct answer is: (A) There is no change in the proton-number and neutron number during gamma-ray emission from a nucleus.

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Most popular questions from this chapter

The total energy of the electron in the first excited state of hydrogen is \(-3.4 \mathrm{eV}\). what is the kinetic energy of the electron in this state? (A) \(6.8 \mathrm{eV}\) (B) \(3.4 \mathrm{eV}\) (C) \(-3.4 \mathrm{eV}\) \((\mathrm{D})-6.8 \mathrm{eV}\)

If \(\mathrm{M}_{0}\) is the mass of an isotope, ${ }^{17}{ }_{8} \mathrm{O}, \mathrm{M}_{\mathrm{p}}\( and \)\mathrm{M}_{\mathrm{n}}$ are the masses of a Proton and neutron respectively, the binding energy of the isotope is (A) \(\left(\mathrm{M}_{0}-8 \mathrm{M}_{\mathrm{p}}\right) \mathrm{C}^{2}\) (B) $\left(\mathrm{M}_{0}-8 \mathrm{M}_{p}-9 \mathrm{M}_{\mathrm{n}}\right) \mathrm{C}^{2}$ (C) \(\left(\mathrm{M}_{0}-17 \mathrm{M}_{\mathrm{n}}\right) \mathrm{C}^{2}\) (D) \(\mathrm{M}_{\mathrm{O}} \mathrm{C}^{2}\)

The masses of neutron and Proton are \(1.0087\) amu and \(1.0073\) amu respectively. It the neuron and Protons combins to form Helium nucleus of mass \(4.0015\) amu then binding energy of the helium nucleus will be (A) \(14.2 \mathrm{MeV}\) (B) \(28.4 \mathrm{MeV}\) (C) \(27.3 \mathrm{MeV}\) (D) \(20.8 \mathrm{MeV}\)

The binding energy Per nucleon of deuteron $\left({ }^{2}{ }_{1} \mathrm{H}\right)\( and Lielium nucleus \){ }_{2}{ }^{4}{ }_{2} \mathrm{He}$ ) is \(1.1 \mathrm{MeV}\) and \(7.0 \mathrm{MeV}\). respectively. If two deuteron react to form a single helium nucleus, the energy released is (A) \(23.6 \mathrm{MeV}\) (B) \(26.9 \mathrm{MeV}\) (C) \(13.9 \mathrm{MeV}\) (D) \(19.2 \mathrm{MeV}\)

The energy released by the fission of one uranium atom is \(200 \mathrm{MeV}\). The number of fission Per second required to Produce \(3.2 \mathrm{w}\) of Power is (A) \(10^{10}\) (B) \(10^{7}\) (C) \(10^{12}\) (D) \(10^{11}\)
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