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CBSE 12th Standard Physics Subject Nuclei HOT Questions 1 Mark Questions 2021

Published on: 27/05/2021

QB365 Provides the HOT Question Papers for Class 12 Physics, and also provide the detail solution for each and every HOT Questions. HOT Questions will help to get more idea about question pattern in every exams and also will help to get more marks in Exams

Download CBSE Class 12th Standard CBSE Physics question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 12th Standard CBSE Physics

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Std: 12th Standard CBSE - Physics

chapter: - HOT Questions

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Questions:

Generic 2

5 x 2 = 10

Two nuclei have mass numbers in the ratio 1:8.What is the ratio of their nuclear radii?

What characteristic property of nuclear force explain the constancy of binding energy per nucleon (BE/A) in the range of mass number 'A' lying 30

In a typical nuclear reaction e.g
\(_{ 1 }^{ 2 }{ H+_{ 1 }^{ 2 }H{ \longrightarrow _{ 2 }^{ 3 }{ He } }+n+3.27MeV }\)
although number of nucleons is conserved,yet energy is released.How?Explain.

In a particular fission reaction, a \(_{ 92 }{ { U }^{ 235 } }\) nucleus captures a slow neutron. The fission products are three neutrons, a \(_{ 57 }{ La^{ 142 } }\)nucleus and a fission product \(_{ Z }{ { X }^{ A } }\). What is the value of Z?

Compare the radii of two nuclei with mass numbers 1 and 27 respectively.

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Answers Key:

Generic 2

5 x 2 = 10

\(R=R_0 A^{1 / 3}, R_1 / R_2=A_1^{1 / 3} / A_2^{1 / 3}=1,{ }^{1 / 3} / 27_{1 / 3}=1 / 3\)

Saturation/Short nature of nuclear force

Since the total initial mass of nuclei on the left side of reaction is greater than the total final mass of nucleus on the right hand side, this difference of mass appears as the energy released.

Applying the principle of conservation of charge number.
Z + 57 = 92
∴ Z = 92 − 57 = 35

\(\frac{R_1}{R_2}=\left(\frac{A_1}{A_2}\right)^{1 / 3}=\left(\frac{1}{27}\right)^{1 / 3}=\frac{1}{3}\)