In a nuclear fission process, a high mass nuclide (A ≈ 236) with binding energy 7.6 MeV/Nucleon dissociated into middle mass nuclides (A ≈ 118), having binding energy of 8.6 MeV/Nucleon. The energy released in the process would be ____ MeV.
Energy released \( Q = \text{BE}_{\text{Product}} - \text{BE}_{\text{Reactant}} \):
\[ Q = 2 \times 118 \times 8.6 - 236 \times 7.6 \]
\[ = 236 \times (8.6 - 7.6) = 236 \, \text{MeV} \]
Match the LIST-I with LIST-II
LIST-I (Type of decay in Radioactivity) | LIST-II (Reason for stability) | ||
---|---|---|---|
A. | Alpha decay | III. | Nucleus is mostly heavier than Pb (Z=82) |
B. | Beta negative decay | IV. | Nucleus has too many neutrons relative to the number of protons |
C. | Gamma decay | I. | Nucleus has excess energy in an excited state |
D. | Positron Emission | II. | Nucleus has too many protons relative to the number of neutrons |
Choose the correct answer from the options given below:
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The density of the copper ($^{64}Cu$) nucleus is greater than that of the carbon ($^{12}C$) nucleus.
Reason (R): The nucleus of mass number A has a radius proportional to $A^{1/3}$.
In the light of the above statements, choose the most appropriate answer from the options given below:
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: