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
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

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.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)