To determine the energy required to remove a neutron from $_{6}^{13}C$, we need to calculate the difference in binding energy between the isotopes $_{6}^{12}C$ and $_{6}^{13}C$. Here are the steps and calculations:
Therefore, the energy required to remove a neutron from $_{6}^{13}C$ is 4.95 MeV.
To remove a neutron from \( ^{13}_6C \), the nuclear reaction can be represented as:
\(^{13}_6C \rightarrow ^{12}_6C + \text{neutron}.\)
The mass defect \(\Delta m\) is given by:
\(\Delta m = \left(12.000000 + 1.008665\right) - 13.003354 = -0.00531 \, \text{u}.\)
The energy required for this process is calculated using:
\(E = \Delta m \times 931.5 \, \text{MeV/u}.\)
Substituting values:
\(E = 0.00531 \times 931.5 \approx 4.95 \, \text{MeV}.\)
The Correct answer is: 4.95 MeV
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 \( C_{t-1} = 28, C_t = 56 \) and \( C_{t+1} = 70 \). Let \( A(4 \cos t, 4 \sin t), B(2 \sin t, -2 \cos t) \text{ and } C(3r - n_1, r^2 - n - 1) \) be the vertices of a triangle ABC, where \( t \) is a parameter. If \( (3x - 1)^2 + (3y)^2 = \alpha \) is the locus of the centroid of triangle ABC, then \( \alpha \) equals: