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Nuclear Structure

ICSE · Class 12 · Physics

Flashcards for Nuclear Structure — ICSE Class 12 Physics. Quick Q&A cards covering key concepts, definitions, and formulas.

63 questions32 flashcards5 concepts

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32 Flashcards
Card 1Composition and size of nucleus

What is the nucleus and how small is it compared with the atom?

Answer

The nucleus is the tiny central part of the atom where almost all the mass and positive charge are concentrated. Its diameter is of the order of 10^-15 m, while the atom size is of the order of 10^-10

Card 2Composition of nucleus

State the main reason why the proton-electron hypothesis for the nucleus was rejected.

Answer

Three main reasons reject it: 1) An electron confined to nuclear size should have energy of about 100 MeV, but beta particle energies are only 2-3 MeV. 2) The magnetic moment of the nucleus is about o

Card 3Neutron importance

Why is the neutron important in nuclear reactions and fission?

Answer

A neutron has no charge, so it is not deflected by the positive nucleus and can enter the nucleus easily. A slow or thermal neutron with energy less than 0.03 eV is more efficient because it spends mo

Card 4Nuclear constituents

State the quark composition of a proton and a neutron.

Answer

A proton contains two up quarks and one down quark. A neutron contains two down quarks and one up quark. The charge on an up quark is +2/3 e and on a down quark is -1/3 e.

Card 5Atomic number and mass number

A nucleus has mass number 23 and atomic number 11. Find the number of neutrons.

Answer

Given: A = 23, Z = 11 Formula: N = A - Z Step 1: N = 23 - 11 Step 2: N = 12 Answer: The nucleus has 12 neutrons.

Card 6Atomic number and mass number

A lithium nucleus has atomic number 3 and mass number 7. How many protons and neutrons does it contain?

Answer

Given: Z = 3, A = 7 Formula: N = A - Z Step 1: Number of protons = Z = 3 Step 2: Number of neutrons = 7 - 3 = 4 Answer: The nucleus contains 3 protons and 4 neutrons.

Card 7Nuclear size

A gold nucleus has mass number 197. Use R = R0 A^(1/3) with R0 = 1.2 x 10^-15 m to find its nuclear radius.

Answer

Given: A = 197, R0 = 1.2 x 10^-15 m Formula: R = R0 A^(1/3) Step 1: 197^(1/3) is about 5.82 Step 2: R = (1.2 x 10^-15 m) x 5.82 Step 3: R ≈ 6.98 x 10^-15 m Answer: The nuclear radius is about 7.0 x 10

Card 8Nuclear size

State the relation between nuclear radius and mass number.

Answer

Nuclear radius: R = R0 A^(1/3) Here, R0 is about 1.2 x 10^-15 m and A is the mass number. This shows that nuclear size increases slowly with mass number.

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