Nuclei MCQs for NEET — Physics Questions with Answers

Energy released per uranium atom = 200 MeV $= 200\times 10^6 ev = 200 \times 10^6 \times 1.6 \times 10 ^{-19}= 320\times10^{-13} \gamma $ $For 320 \times 10^{-13} 2 \rightarrow 1 atom$ $32 J \rightarrow \lambda {3.2 \over 320 \times 10^{-13}} =10^{11}atoms$

The size of a nucleus is generally of the order of $10^{-14}$ meters. This is much smaller than the size of an atom, which is of the order of $10^{-10}$ meters. The nucleus is composed of protons and neutrons and is densely packed.

The ratio of the atomic volume to the nuclear volume is of the order of $10^{15}$. This is because the atomic volume is determined by the electron cloud, which is much larger in size compared to the densely packed nucleus.

$ { volume of atom \over volume of nuclear } = {4\pi \lambda l_1^3 \over 4/ 3 \pi \lambda 1_2^3 } = {(10^{-10})^3 \over (10^{-15})^3 }= 10^{15} $

The nuclei $_7N^{14}$ and $_6C^{13}$ have the same number of neutrons (7). Nuclei with the same number of neutrons but different numbers of protons are called isotones.

To find the fraction of Plutonium remaining after 72000 years, we need to determine how many half-lives have passed. The half-life of Plutonium is 24000 years. After 72000 years, the number of half-lives is:\[ \frac{72000}{24000} = 3 \] The fraction remaining after 3 half-lives is:\[ \left( \frac{1}{2} \right)^3 = \frac{1}{8} \] Therefore, the correct answer is \( \frac{1}{8} \).

A value & Z value $ _{36}Kr ^{89}$

$ _6Cl^2 + _0n^1 \rightarrow N_7^{13} + e_{-1} ^0$

$ E = (\triangle m)C^2$ $ E = (2m-M) C^2$ ${ E \over C^2} = 2m-M$ $ M = 2m - {E/C^2}$

$Ionigation energy of atom E = {13.6 Z^2 \over n^2} ev$ $For helium z = 2= { 13.6 \times (2)^2 \over (1)^1 } = 54.4 ev$