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 small bob A of mass m is attached to a massless rigid rod of length 1 m pivoted at point P and kept at an angle of 60° with vertical. At 1 m below P, bob B is kept on a smooth surface. If bob B just manages to complete the circular path of radius R after being hit elastically by A, then radius R is_______ m :
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 