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The size of the nucleus is of the order of
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 atomic volume of nuclear volume is of the order of
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.
Nucleon is common name for
$ { 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 nucler $_7N^{14}$ and $_6C^{13}$ can be described as
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.
Plutonium decays with half life time 24000 year. if Plutonium is stored after 72000 yes, the fraction of it that remain
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 \( \frac{1}{8} \).
The radio of minimum to maximum wave length in Balmer series is
$ {N \over N0} = ( {1\over 2} ) ^ {t /T1/2} = ({ 1 \over2}) {72000 \over 24000} = {1 \over 8} $
$_6C^{12}$ absorbs an energetic neutron and emits a $\beta$ Particle. The resulting nucleus is
A value & Z value $ _{36}Kr ^{89}$
two deutrons each of mass m fuse to form helium resulting in release of energy E the mass of helium formed is
$ _6Cl^2 + _0n^1 \rightarrow N_7^{13} + e_{-1} ^0$
What percent of original radio active substans is left after 5 half life time
$ E = (\triangle m)C^2$ $ E = (2m-M) C^2$ ${ E \over C^2} = 2m-M$ $ M = 2m - {E/C^2}$
In which region of electro magnetic spectum does the Lyman series of hydrogen atom like
$at \,t = 0 \; \;\; N =No$ $ t = T_ {1/2} \times 1 \;\;\; N = No/2 \;\;\; \therefore {N \over No} = {1 \over 16}$ $ t = T_ {1/2} \times 2 ;;; N = No/4 ;;; {N\over No} \times 100 = { 1 \over 32 } \times 100$ $ t = T_ {1/2} \times 3 \;\;\; N = No/8 \;\;\; =3.125 $ $ t = T_ {1/2} \times 4 \;\;\; N = No/16 \;\;\; \approx 3 \%$ $ t = T_ {1/2} \times 5 \;\;\; N = No/32 $
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